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August 21, 2019November 10, 2019 edeany

Adjunctions

Note: for a category $C$ , we write $[X, Y]_C$ for the hom-class of maps from $X$ to $Y$ .

An adjunction is a pair of functors $F : C \rightarrow D$ and $G : D \rightarrow C$ with a certain relationship. There are many texts expositing adjunctions, but I have not seen one covering what I think of as their fundamental theorem. This is regrettable, since I think many people would appreciate the simplifying effect this result has on the theory of adjunctions in $\text{Cat}$ . My goal today is to cover this theorem using coend calculus.

Adjunctions in a nutshell. First, an example. Let $\text{Set}$ be the category of sets and let $R \text{-mod}$ be the category of $R$ -modules for a ring $R$ . There are functors $F : \text{Set} \rightarrow R \text{-mod}$ sending a set $X$ to $\oplus_{x \in X} R$ and a functor $G : R \text{-mod} \rightarrow \text{Set}$ sending an $R$ -module $M$ to its underlying set. There is a natural way of going back and forth between maps of $R$ -modules $F(X) \rightarrow M$ and maps of sets $X \rightarrow M$ . For a map of $R$ -modules, $f: F(X) \rightarrow M$ , we get a map of sets by first applying $G$ (this really does nothing) and then precomposing with a map $\eta_X$ of sets $X \rightarrow G(F(X))$ sending $x$ to the element $(a_y)_{y \in X}$ in the underlying set of $F(X)$ where $a_x = 1$ and $a_y =0$ for $y \neq x$ . The resulting map is $X \stackrel{\eta_X}{\rightarrow} G(F(X)) \stackrel{G(f)}{\rightarrow} G(M)$ . For a map of sets $f: X \rightarrow G(M)$ , we get a map $F(X) \stackrel{f}{\rightarrow} F(G(M)) \stackrel{\epsilon_M}{\rightarrow} M$ where $\epsilon_M: F(G(M)) \rightarrow M$ sends $(a_x)_{x \in X}$ to $\sum_{x \in X} a_X$ , where the sum is taken in $M$ . What we have here is a natural correspondence between hom sets $[F(X), M]_{R \text{-mod}}$ and $[X, G(M)]_{\text{Set}}$ . To go one way we apply $G$ and then precompose with $\eta_X$ (kind of like correcting for an error, since $F$ and $G$ are not inverses), and to go the other way we apply $F$ and then postcompose with $\epsilon_M$ . This hints at the definition of an adjoint:

Definition: Let $F : C \rightarrow D$ and $G : D \rightarrow C$ be functors. We say that $F$ is adjoint to $G$ , or that $F$ and $G$ are adjoint, if there is a natural isomorphism $\Phi : [ F-, -]_D \rightarrow [-, G-]_C$ between the functors $[ F-, -]_D, [-, G-]_C : C^{op} \times D \rightarrow \text{Set}$ .

The example also hints at what I call the fundamental theorem of adjunctions in the $2$ -category $\text{Cat}$ :

Theorem: Let $C$ and $D$ be categories and $F$ and $G$ be functors. Then
(i) $[[F-, -], [-, G-]]_{[C^{op} \times D, \text{Set}]}$ and $[1_C, GF]_{[C^{op}, C^{op}]}$ are naturally isomorphic.
(ii) $[[-, G-], [F-, -]]_{[C^{op} \times D, \text{Set}]}$ and $[FG, 1_D]_{[D, D]}$ are naturally isomorphic.

Proof: These are dual; we prove the first. For the isomorphism itself, we have a series of natural isomorphisms:

Screen Shot 2019-08-20 at 9.47.23 PM.jpg

This finishes the proof.

Using coend calculus like that is nice, but we are also interested in an explicit description of the isomorphisms $\alpha : \text{Nat}_{[C^{op} \times D, \text{Set}]} ([F-, -], [-, G-]) \rightarrow \text{Nat}_{[C^{op}, C^{op}]}(1_C, GF)$ and $\beta : \text{Nat}_{[C^{op}, C^{op}]}(1_C, GF) \rightarrow \text{Nat}_{[C^{op} \times D, \text{Set}]} ([F-, -], [-, G-])$ ! To achieve this description, notice that the following diagram commutes for each $X \in \text{Obj}(C)$ and each $Y \in \text{Obj}(D)$ :

Screen Shot 2019-08-20 at 9.48.31 PM.jpg

Take a natural transformation $\eta : 1 \rightarrow GF$ . For each $X$ , define a natural transformation $H_{X} : [FX, -] \rightarrow [X, G(-)]$ such that $(H_{X})_Y : [FX, Y] \rightarrow [X, GY]$ sends $f$ to $G(f) \circ \eta_X$ , to $(H_{X})_Y$ . Define a natural transformation $H : [[F(-), -]_D, [-, G(-)]_{C}]_{[C^{op} \times D, \text{Set}]}$ where $H_{X, Y} = (H_X)_Y$ . The composition

Screen Shot 2019-08-20 at 9.27.37 PM

gives the following diagram of mappings:

Screen Shot 2019-08-20 at 10.03.48 PM.jpg

As this holds for each $X \in \text{Obj}(C)$ and each $Y \in \text{Obj}(D)$ , we have $\alpha(\eta) = H$ .

Next take a natural transformation $H : [[F(-), -]_D, [-, G(-)]_{C}]_{[C^{op} \times D, \text{Set}]}$ . For each $X$ , define a natural transformation $H_{X} : [FX, -] \rightarrow [X, G(-)]$ such that $(H_{X})_Y = H_{X, Y}$ . Define a natural transformation $\eta : 1_C \rightarrow GF$ such that $\eta_X = (H_X)_{FX}(1_{FX})$ . The composition

Screen Shot 2019-08-20 at 9.27.37 PM

gives the following diagram of mappings:

Screen Shot 2019-08-20 at 10.04.08 PM.jpg

As this holds for each $X \in \text{Obj}(C)$ and each $Y \in \text{Obj}(D)$ , $\beta(H) = \eta$ .

Hence $\alpha$ sends a natural transformation $\eta : 1_C \rightarrow GF$ to the natural transformation $H : [F-, -] \rightarrow [-, G-]$ where $H_{X, Y}(f)= Gf \circ \eta_X$ for $X \in \text{Obj}(C)$ , $Y \in \text{Obj}(D)$ , and $f : FX \rightarrow Y$ , and $\beta$ sends a natural transformation $H : [F(-), -]_D \rightarrow [-, G(-)]_{C}$ to the natural transformation $\eta : 1_C \rightarrow GF$ such that $\eta_X = H_{X, FX} (1_{FX})$ .

This theorem puts us in a place where we can prove other results about adjoints much more easily:

Theorem: The following are equivalent:

(i) The functors $[F-, -]_D : C^{op} \times D \rightarrow \text{Set}$ and $[-, G-]_C : C^{op} \times D \rightarrow \text{Set}$ are naturally isomorphic.
(ii)(a) (Universal Morphisms Definition) There is a natural transformation $\eta : 1_{C} \rightarrow G F$ such that, for each $X \in \text{Obj}(C)$ , each $Y \in \text{Obj}(D)$ , and each morphism $f : X \rightarrow GY$ in $C$ , there is a unique morphism $g : FX \rightarrow Y$ in $\text{Mor}(D)$ such that $Gg \circ \eta_X =f$ .
(ii)(b) (Universal Morphisms Definition) There is a natural transformation $\epsilon : FG \rightarrow 1_{D}$ such that, for each $Y \in \text{Obj}(D)$ , each $X \in \text{Obj}(C)$ , and each $g : FX \rightarrow Y$ in $D$ , there is a unique morphism $f : X \rightarrow GY$ in $C$ such that $\epsilon_Y \circ Ff = g$ .
(iii) (The Unit Counit Definition) There are natural transformations $\epsilon : FG \rightarrow 1_{D}$ and $\eta : 1_{C} \rightarrow G F$ such that the following compositions are the identity morphism, for each $X \in \text{Obj}(C)$ and $Y \in \text{Obj}(D)$ .

Screen Shot 2019-08-20 at 9.52.01 PM.jpg

Proof: $(i) \implies (ii)(a)$ . Suppose that the functors $[F-, -]_D : C^{op} \times D \rightarrow \text{Set}$ and $[-, G-]_C : C^{op} \times D \rightarrow \text{Set}$ are naturally isomorphic, and take a natural isomorphism $H : [F-, -]_D \rightarrow [-, G-]_C$ . Let $\eta : 1_C \rightarrow GF$ be the corresponding natural transformation. Take $X \in \text{Obj}(C)$ , $Y \in \text{Obj}(D)$ , and $f : X \rightarrow GY$ . There is a unique $g : FX \rightarrow Y$ such that $Hg = f$ , since $H$ is a natural isomorphism. But $Hg = Gg \circ \eta_X$ by the fundamental theorem. This shows (ii)(a).

$(ii)(a) \implies (i)$ . Suppose that there is a natural transformation $\eta : 1_{C} \rightarrow G F$ such that, for each $X \in \text{Obj}(C)$ , each $Y \in \text{Obj}(D)$ , and each morphism $f : X \rightarrow GY$ in $C$ , there is a unique morphism $g : FX \rightarrow Y$ in $\text{Mor}(D)$ such that $Gg \circ \eta_X =f$ . Let $H : [F-, -]_D \rightarrow [-, G-]_C$ be the natural transformation corresponding to $\eta$ . Take $X \in \text{Obj}(C)$ and $Y \in \text{Obj}(D)$ . $H_{X, Y}g = Gg \circ \eta_X$ for each $g : X \rightarrow GY$ , so that, for each $f \in [X, GY]_C$ , there is a unique $g \in [FX, Y]_D$ such that $H_{X, Y}g = Gg \circ \eta_X = f$ . This shows (i).

$(i) \implies (ii)(b)$ . Suppose that the functors $[F-, -]_D : C^{op} \times D \rightarrow \text{Set}$ and $[-, G-]_C : C^{op} \times D \rightarrow \text{Set}$ are naturally isomorphic, and take a natural isomorphism $H : [-, G-]_C \rightarrow [F-, -]_D$ . Let $\epsilon : FG \rightarrow 1_D$ be the corresponding natural transformation. Take $X \in \text{Obj}(C)$ , $Y \in \text{Obj}(D)$ , and $g : FX \rightarrow Y$ . There is a unique $f : X \rightarrow GY$ such that $Hf = g$ , since $H$ is a natural isomorphism. But $Hf = \epsilon_Y \circ Ff$ by the fundamental theorem. This shows (ii)(b).

$(ii)(b) \implies (i)$ . Suppose that there is a natural transformation $\epsilon : FG \rightarrow 1_{D}$ such that, for each $X \in \text{Obj}(C)$ , each $Y \in \text{Obj}(D)$ , and each morphism $g : FX \rightarrow Y$ in $D$ , there is a unique morphism $f : X \rightarrow GY$ in $\text{Mor}(D)$ such that $\epsilon_Y \circ Ff \circ = g$ . Let $H : [-, G-]_C \rightarrow [F-, -]_D$ be the natural transformation corresponding to $\eta$ . Take $X \in \text{Obj}(C)$ and $Y \in \text{Obj}(D)$ . $H_{X, Y}f = \epsilon_Y \circ Ff$ for each $f : X \rightarrow GY$ , so that, for each $g \in [FX, Y]_C$ , there is a unique $f \in [X, GY]_D$ such that $H_{X, Y}f = \epsilon_Y \circ Ff = g$ . This shows (i).

$(i) \implies (iii)$ . Suppose that the functors $[F-, -]_D : C^{op} \times D \rightarrow \text{Set}$ and $[-, G-]_C : C^{op} \times D \rightarrow \text{Set}$ are naturally isomorphic, and take a natural isomorphism $H : [F-, -]_D \rightarrow [-, G-]_C$ . Let $\eta : 1_C \rightarrow GF$ be the natural transformation corresponding to $H$ and let $\epsilon : GF \rightarrow 1_D$ be the natural transformation corresponding to $H^{-1}$ . For $X \in \text{Obj}(C)$ , we have
Screen Shot 2019-08-20 at 9.58.44 PM.jpg
and

$(iii) \implies (i)$ . Suppose $(iii)$ holds. Let $E : [F-, -] \rightarrow [-, G-]$ be the natural transformation induced by $\epsilon : FG \rightarrow 1_D$ , and let $H : [-, G-] \rightarrow [F-, -]$ be the natural transformation induced by $\eta : 1_C \rightarrow GF$ .

Take $Y \in \text{Obj}(D)$ . Write $(EH)_{-, Y} : [-, GY] \rightarrow [-, GY]$ for the induced natural transformation. The following diagram commutes by the Yoneda lemma:
Screen Shot 2019-08-20 at 9.59.32 PM.jpg
So $(EH)_{X, Y} (f) = [f, GY] \circ (EH)_{GY, Y} (1_{GY})= (EH)_{GY, Y}(1_{GY}) \circ f = f$ . So $(EH)_{-, Y}$ is the identity natural transformation. Hence $EH$ is the identity natural transformation.

Take $Y \in \text{Obj}(D)$ . Write $(HE)_{X, -} : [FX, -] \rightarrow [FX, -]$ for the induced natural transformation. The following diagram commutes by the Yoneda lemma:
Screen Shot 2019-08-20 at 10.00.31 PM.jpg
So $(HE)_{X, Y} (f) = [FX, f] \circ (HE)_{X, FX} (1_{FX}) = f \circ (HE)_{X, FX}(1_{FX}) = f$ . So $(HE)_{X,-}$ is the identity natural tarnsformation. Hence $HE$ is the identity natural transformation.

It follows that $H$ and $E$ are inverse natural transformations, and therefore natural isomorphisms. This shows (i).

December 28, 2018August 21, 2019 edeany

The Fundamental Theorem of Galois Theory

This post is the third of three parts, culminating in a proof of the (classical) fundamental theorem of Galois Theory:

1) Finite Separable Algebras
2) Galois Connections
3) The Fundamental Theorem of Galois Theory

We use the same notation as we used in (1) and (2), so read those if you don’t know any of the terms here. For the reader who is looking for the minimal route to a proof of the fundamental theorem, I have put a star (*) next to the essential parts of this post.

Definition:* Let $K/k$ be a finite field extension. We say $K /k$ is Galois if $K$ is a $K$ -separable $k$ -algebra.

Lemma: Let $K/k$ be a finite field extension, and let $\overline{K}$ be the algebraic closure of $K$ . $K/k$ is Galois if and only if $K$ is separable and the canonical map $[K, K]_k \rightarrow [K, \overline{K}]_k$ is surjective.

Proof: If the canonical map above is surjective, then $\# [K, K]_k = \# [ K, \overline{K} ]_k$ , so $\# [K, K]_k = \# [K, \overline{K}]_k = \text{dim}_k (K)$ . Hence $K$ is $K$ -separable.

Conversely, suppose $K$ is $K$ -separable. Then $\text{dim}_k (K) = \# [K, K]_k$ , so $\# [ K, \overline{K} ]_k \geq \# [K, K]_k = \text{dim}_k (K)$ , so that $K$ is separable. Since $\# [K, K ]_k = [K, \overline{K} ]_k$ , the map $[K, K]_k \rightarrow [K, \overline{k}]_k$ is surjective, as desired.

The Fundamental theorem of Galois theory asserts that there is a Galois correspondence between intermediate fields of a Galois field extension $K /k$ and subgroups of $\text{Aut}_k (K)$ . Going along with more recent approaches to the theorem, we choose here to write in terms of group actions instead of subgroups. Here we show that there is a Galois correspondence between intermediate fields of a finite Galois field extension $K/k$ and quotient $G$ -sets of $G$ . This exercise shows that the two approaches are equivalent:

Exercise: there is a preorder isomorphism between subgroups of a group $G$ and quotient $G$ -sets of $G$ . The correspondence sends a subgroup $H$ of $G$ to the quotient $G$ -set $G/H$ and a quotient $G$ -set $X$ with quotient map $\pi : G \rightarrow X$ to $\text{Stab}_{G} (\pi (e))$ , where $e$ is the identity element in $G$ .

Setup: * Let $\text{Sub}(K)$ be the preorder of subobjects of $K$ in the category of $k$ -algebras. Let $\text{Quot}(G)$ be the preorder of quotient objects of $G$ in the category of $G$ -sets.

For a $k$ -algebra $F$ , the set $[F, K]_k$ of $k$ -algebra maps from $F$ to $K$ has the structure of a $G$ -set. $\sigma \in G$ acts on $[F, K]_k$ by sending $\iota : F \rightarrow K$ to $\sigma \circ \iota : F \rightarrow K$ . Define a preorder map $[-, K]_k : \text{Sub}(K) \rightarrow \text{Quot}(G)$ sending a $k$ -algebra $F$ with monomorphism $\iota : F \rightarrow K$ to the $G$ -set $[F, K]_k$ with epimorphism $G \rightarrow [F, K]_k$ sending $\sigma$ to $\sigma \circ \iota$ .

For a $G$ -set $X$ , the set $[X, K]_G$ of $G$ -set maps from $X$ to $K$ has the structure of a $k$ -algebra. For two $G$ -sets $f : X \rightarrow K$ and $g : X \rightarrow K$ , we set $f + g : X \rightarrow K$ to send $x$ to $f(x) + g(x)$ and $fg : X \rightarrow Y$ to send $x$ to $f(x)g(x)$ . For an element $a \in k$ and $f \in [ X, K]_G$ , we set $af : X \rightarrow K$ to send $x$ to $af(x) = f(ax)$ . Define a preorder map $[-, K]_G : \text{Quot}(G) \rightarrow \text{Sub}(K)$ sending a $G$ -set $X$ with epimorphism $\pi : G \rightarrow X$ to the $k$ -algebra $[X, K]_k$ with monomorphism $[X, K]_k \rightarrow K$ sending $\phi$ to $\phi \circ \pi (1)$ .

$[-, K]_k$ and $[-, K]_G$ form a Galois connection. Indeed, taking a subobject $F$ of $K$ and a quotient object $X$ of $G$ , any map $\phi : F \rightarrow [X, K]_G$ of $k$ -algebras gives a map $\psi : X \rightarrow [F, K]_k$ of $G$ -sets sending $x \in X$ to the map $\psi(x) : F \rightarrow K$ of $k$ -algebras sending $a$ to $\phi(a)(x)$ . Conversely, any map $\psi : X \rightarrow [F, K]_k$ of $G$ -sets gives a map $\phi : F \rightarrow [X, K]_G$ of $k$ -algebras sending $a \in F$ to the map $\phi(a) : X \rightarrow K$ of $G$ -sets sending $x$ to $\psi(x)(a)$ .

At last, we have the fundamental theorem of Galois theory:

Theorem: (The Fundamental Theorem of Galois Theory) Let $K/k$ be a finite Galois field extension. Put $G = [K, K]_k$ . There is a preorder isomorphism between $\text{Quot}(G)$ and $\text{Sub}(K)$ given by $[-, K]_k$ and $[-, K]_G$ .

Proof: Take a $k$ -algebra $F \in \text{Sub}(K)$ . Put $\hat{F} = [[F, K]_k, K]_G$ . Then $[\hat{F}, K]_k = [[[F, K]_k, K]_G, K]_k \cong [F, K]_k$ by the main proposition in my post on Galois connections. Hence $\# [\hat{F}, K]_k = \# [F, K]_k$ . $K$ is $K$ -separable since it is Galois (we showed this above). $F$ and $\hat{F}$ are both subobjects of $K$ , and so they are both $K$ -separable, by the stability results in my post on finite separable algebras. So $\text{dim}_k (F) = \# [ F, K]_k$ and $\text{dim}_k (\hat{F}) = \# [ \hat{F}, K]_k$ . It follows that $\text{dim}_k (F) = \text{dim}_k (\hat{F})$ . Now the inclusion $F \rightarrow \hat{F}$ (see my post on Galois connections) is an injection of $k$ -algebras of the same finite dimension, and so must be an isomorphism.

Conversely, take a $G$ -set $X \in \text{Quot}(G)$ . Put $F = [X, K]_G$ , and $\hat{X} = [F, K]_k$ . The quotient map $\pi : G \rightarrow X$ gives an inclusion map $[\pi, K]_G : F \rightarrow K$ . This inclusion tells us that $F$ is $K$ -separable, one of the stability properties in my post on finite separable algebras. So $\text{dim}(F) = \# [F, K ] = \# \hat{X}$ . Also, $\# X \leq \text{dim}(F)$ by part (ii) of the main theorem in my post on finite separable algebras. Now the canonical map $X \rightarrow \hat{X}$ is a surjection and $\# X \leq \# \hat{X}$ , so that it must be an isomorphism of $G$ -sets.

December 27, 2018January 11, 2019 edeany

Galois Connections

This post is the second of three parts, culminating in a proof of the fundamental theorem of Galois Theory:

1) Finite Separable Algebras
2) Galois Connections
3) The Fundamental Theorem of Galois Theory

These posts are designed to aid in the third part, so, for the reader who is looking for the easiest route to a proof of the fundamental theorem, I have put a star (*) next to the essential parts of this post.

Definition:* a preorder is a set $X$ endowed with a relation $\leq$ on $X$ , such that
(i) $x \leq x$ $\forall x \in X$ .
(ii) $x \leq y, y \leq z \implies x \leq z$ $\forall x, y, z \in X$ .

We may view a preorder as a category whose objects are the elements of $X$ , where there is a unique morphism from $x$ to $y$ when $x \leq y$ , and no morphism otherwise. A morphism (functor) of preorders $X$ and $Y$ is a function of sets $f: X \rightarrow Y$ such that $x \leq y$ implies $f(x) \leq f(y)$ . We may also define a contravariant preorder map as a function of sets $f : X \rightarrow Y$ such that $x \leq y$ implies $f(y) \leq f(x)$ . In lieu of the viewpoint that preorders are particular kinds of categories, two elements $x, y \in X$ are called isomorphic when $x \leq y$ and $y \leq x$ .

Definition:* let $P$ and $Q$ be preorders and $f : P \rightarrow Q$ and $g: Q \rightarrow P$ contravariant preorder maps. We say that $f$ and $g$ form a Galois connection if $f$ is left adjoint to $g$ . That is, for each $x \in P$ and each $y \in Q$ , $y \leq f(x)$ if and only if $x \leq g(y)$ . We say that the Galois connection is a Galois correspondence if the the units of the adjunction are natural isomorphisms, in which case $f$ and $g$ testify to a categorical equivalence between the preorders $P$ and $Q$ . Note that, due to the contravariance of the functors $f$ and $g$ , the adjunction has two units, and no counits.

Theorem:* let $P$ and $Q$ be preorders and let $f : P \rightarrow Q$ and $g : Q \rightarrow P$ be contravariant preorder maps forming a Galois connection. Then
(i) $x \leq gf (x) \forall x \in P$ . (N.B. this is the unit map of the adjunction).
(ii) $y \leq fg (y) \forall y \in Q$ . (N.B. this is the counit map of the adjunction).
(iii) $f(x)$ and $fgf(x)$ are isomorphic for all $x \in P$ .
(iv) $g(x)$ and $gfg (y)$ are isomorphic for all $y \in Q$ .
(v) $f|_{\text{im}(g)} : \text{im}(g) \rightarrow \text{im}(f)$ and $g|_{\text{im}(f)} : \text{im}(f) \rightarrow \text{im}(g)$ form a Galois correspondence.

Proof:
(i) For each $x \in P$ , $x \leq gf(x)$ if and only if $f(x) \leq f(x)$ , which is true.
(ii) For each $y \in Q$ , $y \leq fg(y)$ if and only if $g(y) \leq g(y)$ , which is true.
(iii) Take $x \in P$ . Put $y = f(x)$ . $fg(y) \geq y$ by (ii), so $fgf(x) \geq f(x)$ . $x \leq gf(x)$ by (i), so that $fgf(x) \leq f(x)$ . Hence $fgf(x) = f(x)$ .
(iv) Take $y \in Q$ . Put $x = g(y)$ . $gf(x) \geq y$ by (ii), so $gfg(y) \geq g(y)$ . $y \leq fg(y)$ by (i), so that $gfg(y) \leq g(y)$ . Hence $gfg(y) = g(y)$ .
(v) By (iii), $y \leq f|_{\text{im}(g)} \circ g|_{\text{im}(f)} (y) \leq y$ for each $y \in Q$ , and by (iv), $x \rightarrow g|_{\text{im}(g)} \circ f|_{\text{im}(g)}(x) \leq x$ for each $x \in P$ . The claim follows.

N.B. (iii) and (iv) follow from the triangle identities for a (contravariant) adjunction between categories. For preorders, we see that we get a full isomorphism.

Example: let $X$ and $Y$ be sets and let $R \subset X \times Y$ be a relation. Let $P$ be the set of subsets of $X$ , partially ordered by inclusion, and let $Q$ be the set of subsets of $Y$ , partially ordered by inclusion. Define a preorder map $f : P \rightarrow Q$ where $A \subset X$ is sent to the subset $\{ y \in Y : a R y\ \forall\ a \in A \}$ and a preorder map $g : Q \rightarrow P$ where $B \subset Y$ is sent to the subset $\{ x \in X : x R b\ \forall b \in B \}$ . $f$ and $g$ form a Galois connection. Indeed, for subsets $A \subset X$ and $B \subset Y$ , $B \subset f(A)$ if and only if $\forall a \in A \forall b \in B : aRb$ , if and only if $A \subset g(B)$ .

Example: let $k$ be a field, $k[x_1, ..., x_n]$ the polynomial ring over $k$ in $n$ variables. Write $\mathbb{A}^n$ for $k^n$ , $n$ dimensional affine space. Let $P$ be the set of subsets of $k[x_1, ..., x_n]$ and $Q$ be the set of subset of $\mathbb{A}^n$ . Define a relation $R \subset k[x_1, ..., x_n] \times \mathbb{A}^n$ where, for $f(x_1, ..., x_n) \in k[x_1, ..., x_n]$ and $(a_1, ..., a_n) \in \mathbb{A}^n$ , $f(x_1, ..., x_n) R (a_1, ..., a_n)$ if $f(a_1, ..., a_n) = 0$ . Define a map $V : P \rightarrow Q$ sending a subset $X$ of $k[x_1, ..., x_n]$ to the set $V(X) = \{ (a_1, ..., a_n) \in \mathbb{A}^n : f(a_1, ..., a_n) \forall f \in k[x_1, ..., x_n] \}$ and a map $I : Q \rightarrow P$ sending a subset $Y$ of $\mathbb{A}^n$ to the set $I(Y) = \{ f(x_1, ..., x_n) \in k[x_1, ..., x_n] : f(a_1, ..., a_n) \forall (a_1, ..., a_n) \in Y \}$ . Then $f$ and $g$ form a Galois connection. In fact, this is an instance of the above example.

By the proposition above, $I$ and $V$ restrict to inclusion reversing bijections $I|_{\text{im} (V)} : \text{im} (V) \rightarrow \text{im}(I)$ and $V|_{\text{im}(I)} : \text{im}(I) \rightarrow \text{im}(V)$ . When $k$ is algebraically closed, the Nullstellensatz characterizes the image of $I$ as the radical ideals. The subsets of $X$ contained in the image of $V : P \rightarrow Q$ form the closed sets of a topology on $X$ , called the Zariski topology.

Example: $k$ be a field and let $V$ be a $k$ -vector space. Let $P$ be the set of subsets of $V$ , ordered by inclusion. Let $Q$ be the set of subsets of $V^*$ , the dual of $V$ as a $k$ -vector space. There is a relation $R \subset V \times V^*$ where $f R x$ when $f(x) = 0$ . Define a map $I : P \rightarrow Q$ sending a subset $X$ of $V$ to the set $\{ f \in V^* : f(x) = 0 \forall x \in X \}$ , and a map $V : Q \rightarrow P$ sending a subset $Y$ of $V^*$ to the set $\{ x \in V : f(x) = 0 \forall f \in Y \}$ . This also follows from the first example.

As always, $I$ and $V$ restrict to inclusion reversing bijections $I|_{\text{im} (V)} : \text{im} (V) \rightarrow \text{im}(I)$ and $V|_{\text{im}(I)} : \text{im}(I) \rightarrow \text{im}(V)$ . We may characterize the image $\text{im}(V)$ as the $k$ -vector subspaces of $V$ . The elements of the image $\text{im}(I)$ form the closed subsets of a topological space.

Example: Let $M$ be an $R$ -module, and form the ring $S = \text{End}_R (M)$ of endomorphisms of $M$ . Let $P$ be the set of subsets of $S$ . Define a relation $R \subset S \times S$ where $f R g$ when $f \circ g = g \circ f$ . Define a map $I : P \rightarrow P$ sending $X \subset S$ to the set $\{ f \in S : f \circ g = g \circ f \forall g \in X \}$ . $I$ restricts to an inclusion reversing bijection $I|_{\text{im} (I)} : \text{im}(I) \rightarrow \text{im} (I)$ , which is is its own inverse. The set $I(X)$ is called the commutant of $X$ .

Exercise: Let $C$ be a category with a zero object, kernels, and cokernels. Take an object $X$ in $C$ . Let $\text{Sub}(X)$ be the category of subobjects of $X$ and let $\text{Quot}(X)$ be the category of quotient objects of $X$ . These are each preorders. Define a preorder map $\text{cok} : \text{Sub}(X) \rightarrow \text{Quot}(X)$ sending $\iota : Y \rightarrow X$ to $\text{cok}(\iota)$ and a preorder map $\text{ker} : \text{Quot}(X) \rightarrow \text{Sub}(X)$ sending $\pi: X \rightarrow Y$ to $\text{ker}(\pi)$ . Show that this forms a Galois connection.

How could a Galois connection be called such without the namesake example- a Galois connection between subfields of a field and subgroups of its Galois group? The reader may well object that this example was left out. You won’t have to lament for long, though, since this correspondence is the subject of my next post!

December 27, 2018August 21, 2019 edeany

Separable Algebras

This post will be the first of three parts, culminating in a proof of the fundamental theorem of Galois Theory:

1) Finite Separable Algebras
2) Galois Connections
3) The Fundamental Theorem of Galois Theory

The (elementary) fundamental theorem of Galois theory asserts a correspondence between sub-extensions of certain field extensions $K /k$ and subgroups of $\text{Aut}_k (K)$ . Those field extensions for which the theorem holds are appropriately named the Galois field extensions. It is this theorem that motivates me to make this post, but in particular I am motivated because I want to show that this theorem is simpler and altogether more manifest than many expositions would lead you to believe. In point of fact, there are only two challenging theorems at play here, depending on where certain lines are to be drawn. One of these is the fundamental theorem of Galois theory, and another is some equivalent of Artin’s linear independence of characters. In spite of this, some expositions of this fundamental theorem in its most elementary form make it appear to be more tricky than it is. For instance, some authors use primitive element theorem where it isn’t necessary.

In this first post, though, we talk about finite separable algebras. We’ll define the concept of a finite $K$ -separable $k$ -algebra. Every finite $K$ -separable $k$ -algebra will be finite $\overline{k}$ -separable, and we’ll call a $k$ -algebra separable if it is $\overline{k}$ -separable.

Throughout, we set $k$ as a field and $K$ as a finite field extension of $k$ . For $k$ -algebras $A$ and $B$ , we write $[A, B]_k$ for the set of $k$ -algebra maps from $A$ to $B$ . We write $\overline{k}$ for the algebraic closure of $k$ . Throughout, we set $G = \text{Aut}_k (K)$ . For a group $G$ and $G$ -sets $X$ and $Y$ , we write $[X, Y]_G$ for the set of $G$ -set maps from $X$ to $Y$ .

We write $K[G]$ for the group algebra over $G$ . For a $G$ -action $X$ , write $\oplus_{X} K$ for the $K[G]$ -module consisting of formal sums $\sum_{x \in X} a_x x$ where all but finitely many of $a_x$ are zero. The structure map $K[G] \otimes_K \left( \oplus_{X} K \right) \rightarrow \oplus_X K$ is determined by specifying that $g \cdot 1 x = 1 gx$ . Write $\prod_X K$ for the $K[G]$ -module consisting of formal sums $\sum_{x \in X} a_x x$ viewed as functions $X \rightarrow K$ . The structure map $K[G] \otimes_K \left( \prod_{X} K \right) \rightarrow \prod_X K$ is determined by sending $1g \otimes \phi$ to the function $X \rightarrow K$ sending $x$ to $\phi(gx)$ .

Our first result plays the role of Artin’s linear independence of characters in his rendition of Galois theory. The interested reader might examine the proof of (i) below to look for a commonality with Artin’s proof.

Theorem:
(i) For each $k$ -algebra $A$ , there is a canonical map of $k[G]$ -modules, $\epsilon_A : \oplus_{[A, K]} K \rightarrow \text{Hom}_{L \text{-vect}} (A, K)$ which sends $\sum_{\sigma \in [A, L]} a_{\sigma} \sigma$ to the map $\phi : A \rightarrow L$ sending $x$ to $\sum_{\sigma \in [A,K]} a_{\sigma} \sigma(x)$ . $\epsilon_A$ is an injection.
(ii)* For each $G$ -set $X$ , there is a canonical map of $k[G]$ -modules, $\epsilon_X : \oplus_{X} K \rightarrow \text{Hom}_{L \text{-vect}} ([X, L]_G , L)$ which sends $\sum_{x \in X} a_{x} x$ to the map $\phi : [X, L]_G \rightarrow L$ sending $\sigma$ to $\sum_{x \in X} a_{x} \sigma(x)$ . $\epsilon_X$ is an injection.
(iii)* For each $k$ -algebra $A$ , there is a canonical map of $k[G]$ -modules, $\eta_A : K \otimes_k A \rightarrow \prod_{ [A, K]_k} K$ which sends $a \otimes b$ to $(a \sigma(b))_{\sigma \in [A, K]_k}$ . $\eta_A$ is a surjection.
(iv) For each $G$ -set $X$ , there is a canonical map of $k[G]$ -modules, $\eta_X : K \otimes_k [X, K]_G \rightarrow \prod_{X} K$ which sends $a \otimes b$ to $( a b(x) )_{x \in X}$ . $\eta_X$ is a surjection.

Proof:
(i) Take $\sum_{\sigma \in X} a_{\sigma} \sigma$ in the kernel of $\epsilon_A$ , such that the amount of nonzero $a_{\sigma}$ is the least possible. Take $\tau \in X$ , and take $y \in A$ such that $\tau(y) \neq \sigma(y)$ for some $\sigma \in X$ with $a_{\sigma} \neq 0$ . Then $\sum_{i = 1} a_{\sigma} \sigma(y) \sigma (x) = \sum_{i = 1}^n a_{\sigma} \sigma (yx) = 0$ for each $x \in A$ . And $\sum_{i = 1}^n a_{\sigma} \tau(y) \sigma(x) = 0$ , so $\sum_{i = 1}^n (a_{\sigma} \tau(y) - a_{\sigma} \sigma(y)) \sigma$ is contained in the kernel of $\phi$ . Yet this element is nonzero, as $\tau$ and $y$ were chosen so that $\tau(y) - \sigma(y) \neq 0$ for some $\tau \in X$ , and $a_{\sigma} \neq 0$ . So we have a nonzero element of the kernel of $\epsilon_A$ with strictly fewer nonzero summands, a contradiction.
(ii) Take $\sum_{\sigma \in X} a_{\sigma} \sigma$ in the kernel of $\epsilon_A$ , such that the amount of nonzero $a_{\sigma}$ is the least possible. Choose $\tau \in X$ and $\hat{x} \in [X, L]_G$ such that $\hat{x}(\tau) \neq \hat{x}(\sigma)$ and $a_{\sigma} \neq 0$ . Then $\sum_{i = 1} a_{\sigma} \hat{y} (\sigma) \hat{x} (\sigma ) = \sum_{i = 1}^n a_{\sigma} (\hat{y} \hat{x}) ( \sigma ) = 0$ for each $\hat{y} \in A$ . And $\sum_{i = 1}^n a_{\sigma} \hat{y}(\tau) \hat{x}(\sigma) = 0$ , so $\sum_{i = 1}^n (a_{\sigma} \hat{y} (\tau) - a_{\sigma} \hat{y} (\sigma)) \sigma$ is contained in the kernel of $\epsilon_X$ . Yet this element is nonzero, as $\tau$ and $\hat{y}$ were chosen so that $\hat{y}(\tau) - \hat{y}(\sigma) \neq 0$ for some $\tau \in X$ , and $a_{\sigma} \neq 0$ . So we have a nonzero element of the kernel of $\epsilon_A$ with strictly fewer nonzero summands, a contradiction.
(iii) For each $\sigma \in [A, K]_k$ , there is a map $\eta_{A, \sigma} : K \otimes_k A \rightarrow K$ sending $a \otimes b$ to $a \sigma(b)$ with kernel $\mathfrak{m}_i$ . $\mathfrak{m}_i$ are coprime ideals as they are distinct maximal ideals, so they produce a surjective canonical map $\eta_{A} : K \otimes_k A \rightarrow \prod_{ [A, K]_k} K$ by the Chinese remainder theorem.
(iv) For each $\sigma \in X$ , there is a map $\eta_{X, \sigma} : K \otimes [X, K]_G \rightarrow K$ sending $a \otimes \hat{b}$ to $a \hat{b} (\sigma)$ with kernel $\mathfrak{m}_i$ . $\mathfrak{m}_i$ are coprime ideals as they are distinct maximal ideals, so they produce a surjective canonical map $\eta_A : K \otimes_k [X, K]_G \rightarrow \prod_{X} K$ by the Chinese remainder theorem.

Corollary: For each finite dimensional $k$ -algebra $A$ , we have $\#[A, K] \leq \dim_k(A)$ . Then $\#[A, K] = \dim_k(A)$ if and only if $\eta_A$ is an isomorphism.

Proof: $\eta_A$ is a surjection, so that $\eta_A$ is an isomorphism if and only if $\text{dim}_{K}(K \otimes_k A) = \text{dim}_{K}(\prod_{[A, K]} K)$ . But $\text{dim}_{K}(K \otimes_k A) = \text{dim}_k(A)$ and $\text{dim}_{K}(\prod_{ [A, K]}K) = \#[A, K]$ , so $\eta_A$ is an isomorphism if and only if $\#[A, K] = \dim_k(A)$ .

This sets up the stage for our primary definition, and topic of the hour: finite separable $k$ -algebras. If $\#[A, K] \leq \dim_k(A)$ always, then when do we have $\#[A, K] = \dim_k(A)$ ? We start by putting a name on this condition:

Definition:* We say a finite dimensional algebra $A$ is $K$ -separable if $\dim_k (A) = \# [A, K]_k$ . We say that $A$ is separable if it is $\overline{k}$ -separable (we get the same number if we use the separable closure, if you know what that is).

I refer to $\# [A, K]_k$ as the ‘geometric dimension’ of $A$ . Then $A$ being separable says that the geometric dimension of $A$ is the dimension of $A$ .

Proposition: Let $k$ be a field and $K$ a field extension of $k$ . A $k$ -algebra $A$ is $K$ -separable if and only if $K \otimes_k A$ is a free $K$ -module of dimension $\text{dim}_{k}(A)$ .

Proof: first suppose that $K \otimes A$ is a free $K$ -algebra of dimension $\text{dim}_{k}(A)$ . Then there is an isomorphism $K^{\text{dim}_k(A)} \cong K \otimes A$ . $\# [A, K]_k = \# [ K \otimes A, K]_{K} = \# [K^{\text{dim}_k(A)}, K] = \text{dim}_k(A)$ . (That $\# [K^{\text{dim}_k(A)}, K] = \text{dim}_k(A)$ is left as an exercise). So that $A$ is separable. If $A$ is separable, then $\eta_A$ is an isomorphism, so that $K \otimes_k A$ is a free $K$ -algebra of dimension $\text{dim}_k(A)$ .

Lastly, I wanted to show some stability results for finite $K$ -separable $k$ -algebras

Proposition: Let $A$ and $B$ be finite $k$ -algebras. We have the following stability results for $K$ -separability:
(i) If $A$ and $B$ are finite $K$ -separable $k$ -algebras, then $A \otimes_k B$ is finite $K$ -separable.
(ii) If $A$ and $B$ are finite $K$ -separable $k$ -algebras, then $A \prod B$ is finite $K$ -separable.
(iii)* If $A$ is finite $K$ -separable and there exists an injective map $\iota : B \rightarrow A$ , then $B$ is finite $K$ -separable.
(iv) If $A$ is finite $K$ -separable and there exists a quotient map $\pi : A \rightarrow B$ , then $B$ is finite $K$ -separable.
(v) If $A$ is generated by finitely many finite $K$ -separable subalgebras $A_i$ , then $A$ is $K$ -separable.
(vi) $0$ and $k$ are finite $K$ -separable.

Proof:
(i) Suppose $A$ and $B$ are finite $K$ -separable $k$ -algebras. The functor $K \otimes_k - : k \text{-vect} \rightarrow K \text{-vect}$ preserves coproducts. In fact it preserves all colimits, as it is left adjoint (proof of this, in whatever prefered generality, is left to the reader). The upshot is that the canonical algebra map $K \otimes_k (A \otimes_k B) \rightarrow (K \otimes_k A) \otimes_{K} (K \otimes_k B)$ is an isomorphism. Take isomorphisms $K \otimes_k A \cong K^{\text{dim}_k(A)}$ and $K \otimes_k B \cong K^{\text{dim}_k(B)}$ . We then have
Screen Shot 2018-12-27 at 3.02.07 AM.jpg
Thus $K \otimes_k (A \otimes_k B)$ is a free $K$ -algebra of dimension $\text{dim}_k(A \otimes_k B)$ , so that $A \otimes_k B$ is separable.
(ii) Suppose $A$ and $B$ are finite $K$ -separable $k$ -algebras. We may make a similar argument to before to show that the canonical map $K \otimes_k (A \prod B) \rightarrow (K \otimes_k A) \prod (K \otimes_k B)$ is an isomorphism. Take isomorphisms $K \otimes_k A \cong K^{\text{dim}_k(A)}$ and $K \otimes_k B \cong K^{\text{dim}_k(B)}$ . We then have
Screen Shot 2018-12-27 at 3.03.58 AM.jpg
Thus $K \otimes_k (A \prod B)$ is a free $K$ -algebra of dimension $\text{dim}_k(A \prod B)$ , so that $A \prod B$ is separable.
(iii)* Suppose $A$ is finite $K$ -separable and take an injective map $\iota : B \rightarrow A$ of $k$ -algebras. The following diagram commutes:
Screen Shot 2018-12-27 at 3.04.22 AM.jpg
$\text{id}_{K} \otimes_k \iota$ is injective since $L$ is a flat $k$ -module; all $k$ -modules are flat since $k$ is a field. From this it follows that the map $K \otimes_k B \rightarrow \prod_{[B, K]} K$ is injective, so that $B$ is separable by the theorem above.
(iv) Suppose $A$ is finite $K$ -separable and take a surjective map $\pi : A \rightarrow B$ . To show that $B$ is $K$ -separable, we show that $K \otimes_k B$ is a free $K$ -algebra of dimension $\text{dim}_k(B)$ . The composition $K^{\text{dim}_k(A)} \rightarrow K \otimes_k A \rightarrow K \otimes_k B$ is a surjection. Take $n \in \mathbb{N}_{\geq 1}$ to be the $K$ -dimension of the kernel of this map, so that $K^{\text{dim}_k(A)- n} \cong B$ . We have
Screen Shot 2018-12-27 at 3.27.21 AM.jpg
so $\text{dim}_k(B) \text{dim}_k (K) = (\text{dim}_k(A) - n) \text{dim}_k(K)$ , so $\text{dim}_k (A) - n = \text{dim}_k(B)$ .
(v) It suffices to show that, if $A$ is a $k$ -algebra generated by $K$ -separable $k$ -subalgebras $A_1$ and $A_2$ , then $A$ is $K$ -separable. In this case, there is a surjective canonical map $A_1 \otimes_k A_1 \rightarrow A$ sending $a \otimes b$ to $ab$ . $A_1 \otimes_k A_2$ is $K$ -separable by (i), and by (iv) this implies that $A$ is separable.
(vi) This is clear.

February 4, 2018February 4, 2018 edeany

Semi-Direct Products

Let’s talk about semi-direct products! We’ll talk about the categorical significance of semi-direct products and some simple examples. But first, just what is a semi-direct product?

Take a group $G$ with a structure-respecting group action of $G$ on a group $H$ . In other words, we have a homomorphism $\phi : G \rightarrow \text{Aut}(H)$ . Contrast this with a group action of $G$ on $H$ which does not respect the group structure of $H$ , or in other terms a homomorphism $G \rightarrow \text{Bij}(H)$ . From $\phi$ we can form a group $H \rtimes_{\phi} G$ . We write $H \rtimes G$ when the context is clear. As a set this is $H \times G$ , but we put an operation on it which is distinct from the usual group product. Instead, we define $(h, g) (h', g') = (h \phi(g)(h'), g g')$ . This group is called the semi-direct product of $G$ and $H$ . Note that it is dependent on the homomorphism $\phi$ . If $G$ acts on $H$ trivially, then we get the ordinary direct product $G \times H$ , since $(h, g)(h', g') = (h \phi(g)(h') , g g') = (h h', g g')$ . But in general, the semi-direct product is ‘twisted’ in a way, and certainly not isomorphic to $G \times H$ .

Let’s check that this is a group. To see the operation is associative, take $g, g', g'' \in G$ and $h, h', h'' \in H$ . Then
$((h , g) (h', g'))(h'', g'') = (h \phi(g)(h'), g g') (h'', g'') = (h \phi(g)(h') \phi(g g') (h'') , g g' g'' )$
and
$(h , g) ((h', g')(h'', g'')) = (h, g) (h'' \phi(g')(h''), g' g'') = (h \phi(g) (h'' \phi(g')(h'')), g g' g'')$
And the RHS of the two equations is seen to be equal since $\phi$ is a homomorphism and $\phi(g)$ is a homomorphism. $(1, 1)$ is an identity for $H \rtimes G$ : $(1, 1) (h, g) = (\phi(1)(h), g) = (h, g)$ and $(h, g)(1, 1) = (h, g)$ . And lastly we need an inverse for $(h, g)$ . It will have to be of the form $(h, g)(x, g^{-1})$ for some $x \in H$ . We need $h \phi(g)(x) = e$ , so that we can take $x = \phi(g^{-1})(h^{-1})$ and that will do. But we should check that $(x, g^{-1})(h, g) = (1, 1)$ : $(x, g^{-1})(h, g) = (x \phi(g^{-1})(h), 1) = (\phi(g^{-1})(h^{-1}) \phi(g^{-1})(h), 1) = (1, 1)$ . So $H \rtimes G$ is a group.

Let’s look at a few examples. Take an abelian group $A$ . $\mathbb{Z} / 2 \mathbb{Z}$ acts on $A$ by $\overline{1} \cdot a = a^{-1}$ and $\overline{0} \cdot a = a$ . We need to require that $A$ be abelian, so that $\overline{1} \cdot (ab) =(ab)^{-1} = b^{-1} a^{-1} = a^{-1} b^{-1} = (\overline{1} \cdot a) (\overline{1} \cdot b)$ . Under this action, we get a group $A \rtimes \mathbb{Z} / 2 \mathbb{Z}$ . We can describe it with generators and relations as $(A * \mathbb{Z} / 2 \mathbb{Z}) / K$ where $K$ is the subgroup generated by the relations $\{ at = ta^{-1} : a \in A \}$ . When $A = \mathbb{Z} /n \mathbb{Z}$ , we get $D_{n}$ .

Another interesting example: $\mathbb{R}^n$ and $O_n$ . For $x \in \mathbb{R}^n$ and $M \in O_n$ , put $M \cdot x = Mx$ . $\mathbb{R}^n \rtimes O_n \cong E(n)$ , the Euclidian group.

It is not hard to see that $H \rtimes G \cong (H * G)/ K$ where $K$ is the smallest normal subgroup containing the relations $g h = \phi(g) (h) g$ . Take the homomorphism $H * G \rightarrow H \rtimes G$ . It’s surjective, and it can be seen to have kernel $K$ . This gives us a way of expressing the semi-direct product as a colimit, but it’s not the nicest categorical expression at hand. To see the nicer property, let $G \uparrow \text{Grp}$ be the category of groups under $G$ . It’s objects are homomorphisms $\phi : G \rightarrow H$ , and its morphisms $\phi \rightarrow \psi$ are morphisms $\alpha : \text{Cod} (\phi) \rightarrow \text{Cod}(\psi)$ such that $\alpha \circ \phi = \psi$ . We also have the category of $G$ -groups, $G \text{-Grp}$ . Its objects are groups $H$ with morphisms $G \rightarrow \text{Aut}(H)$ , and its morphisms are $G$ -equivariant homomorphisms.

We can think of the objects in $G \uparrow \text{Grp}$ as special types of $G$ -groups. Every morphism $\phi: G \rightarrow H$ induces a $G$ -group, which consists of $H$ with the $G$ -action $g \cdot h = \phi(g) h \phi(g)^{-1}$ . If we replace an object $\phi : G \rightarrow H$ under $G$ with its corresponding $G \text{-Grp}$ $\psi : G \rightarrow \text{Aut}(H)$ , then we lose information about the homomorphism, and it can’t be recovered. But there is a “best approximation”, namely $G \rightarrow H \rtimes_{\psi} G$ !

In categorical terms, we have a forgetful functor $F : G \uparrow \text{Grp} \rightarrow G \text{-Grp}$ whose left adjoint $- \rtimes G : G \text{-Grp} \rightarrow G \uparrow \text{Grp}$ sends objects $\phi : G \rightarrow \text{Aut}(H)$ to objects $H \rtimes_{\phi} G$ . A morphism $H \rightarrow H'$ under $G$ is sent to that same morphism in $G \text{-Grp}$ , and a $G$ -equivariant homomorphism $\alpha : H \rightarrow H'$ is sent to the morphism $\alpha : H \rtimes G \rightarrow H' \rtimes G$ where $(h, g) \mapsto (\alpha(h), g)$ .

To see that this is actually an adjoint relationship, take a $G$ -group $H$ and set a map $\eta_H : H \rightarrow H \rtimes G$ where $h \mapsto (h, 1)$ . We’ll check the unit definition of an adjunction: for each object $G \stackrel{\phi}{\rightarrow} K$ in $G \uparrow \text{Grp}$ and each morphism $\alpha : H \rightarrow F(K)$ , there is a unique morphism $\beta: H \rtimes G \rightarrow K$ such that $F(\beta) \circ \eta_H = \alpha$ .
Screen Shot 2018-02-03 at 8.50.13 PM
Take an object $G \stackrel{\phi}{\rightarrow} K$ and a $G$ -equivariant homomorphism $\alpha : H \rightarrow F(K)$ . We must set $\beta : H \rtimes G \rightarrow K$ where $\beta(h, 1) = \alpha(h)$ and $\beta(1, g) = \phi(g)$ . This gives $\beta(h, g) = \alpha(h, 1)\phi(1, g)$ . Taking $h, h' \in H$ and $g, g' \in G$ . Then
$\alpha(h \phi(g)(h')) \phi(gg') = \alpha( h) \alpha(\phi(g)(h')) \phi(g) \phi(g') = \alpha (h) \phi(g) \alpha(h') \phi(g)^{-1} \phi(g) \phi(g') = \alpha(h) \phi(g) \alpha(h') \alpha(h') \phi(g')$
As desired.

There is a nice characterization of which objects $G \rightarrow H$ in $G \uparrow H$ are isomorphic in $G \uparrow \text{Grp}$ to $G \rightarrow K \rtimes G$ for some $G$ -group $K$ . Recall that retract of a morphism $G \rightarrow H$ is a morphism $H \rightarrow G$ such that $G \rightarrow H \rightarrow G$ is the identity. A morphism $G \rightarrow H$ is isomorphic in $G \uparrow \text{Grp}$ to $G \rightarrow K \rtimes G$ for some $G$ -group $K$ if and only if it has a retract. For a $G$ -group $K$ , it is clear that the composition $G \stackrel{\phi}{ \rightarrow} K \rtimes G \stackrel{\psi}{\rightarrow} G$ , where $\phi(g)= (1, g)$ , and $\psi(1, g) = g$ , is the identity. Conversely, suppose we have a morphism $G \rightarrow H$ with retract $\phi : H \rightarrow G$ . Let $K = \ker(\phi)$ . Since $K$ is normal, the $G$ -group on $H$ where $h \mapsto \phi(g) h \phi(g)^{-1}$ induces a $G$ -group on $K$ . The following diagram commutes, where $\iota : K \rightarrow F(H)$ is the inclusion map: Screen Shot 2018-02-04 at 1.54.42 PM
Note that $\iota$ is $G$ -equivariant by the construction of the action of $G$ on $K$ . Applying the adjoint correspondence, we get a commutative diagram in $G \uparrow \text{Grp}$ :
Examination of the rows reveals that they are in fact exact sequences. That $\alpha$ in the above diagram is an isomorphism follows from the 5-lemma. N.B. the 5-lemma is usually applied in the context of abelian groups or $R$ -modules – or an abelian category, using the Mitchell’s embedding theorem. One can check, however, that the usual diagram chase works here, however. Thus we have an isomorphism $K \rtimes G \rightarrow H$ in $G \uparrow \text{Grp}$ .

Exercise: This observation also allows for a nice internal characterization of a semi-direct product. Suppose that $G$ and $N$ are subgroups of a group $H$ such that (i) $NG = H$ , (ii) $N \cap G = \{ e \}$ , and (iii) $N$ is normal. Show that $H / N \cong G$ , and that there is a section $H / N \rightarrow H$ . What about the converse?