Fun With Representable Functors, or Why I Like Yondea's Lemma.

Sometimes people ask, “why do you use Category Theorey? Isn't it just a set of abstract terms making things look more complicated?” Well, sometimes this is true. Often, it allows to make short and precise statements instead of listing several properties:

“ is a functor from the category of groups into the category of sets” (or “ $F : \catGrp \to \catSet$ is a functor”)

versus

“for every group, , is a set; if $\phi : G \to G'$ is a group homomorphism, then $F(\phi) : F(G) \to F(G')$ is a map; if $\phi = \id_G$ , then $F(\phi) = \id_{F(G)}$ ; if $\psi : G' \to G''$ is another group homomorphism, then $F(\psi \circ \phi) = F(\psi) \circ F(\phi)$ ”.

And it allows to apply very generic statements to a large class of specific examples. And, sometimes, it even gives new insights by abstracting results.

Today I want to discuss representable functors and Yondea's lemma which is, for example, used a lot in modern Algebraic Geometry. In the following, I will always assume that in all categories I use, the morphisms between two objects form a set. We will denote the category of sets by $\catSet$ (and not the french $\catEns$ ), the category of groups by $\catGrp$ , the category of abelian groups by $\catAb$ , the category of rings by $\catRing$ , the category of field extensions of by $\catFld(K)$ and, for a ring , the category of -modules by $\catMod(R)$ . We begin with the definition of a representable functor. For any category $\schmC$ whose objects are sets with further structures, and whose morphisms are maps between these sets, we call $\schmC$ concrete and we have the forgetful functor $Forget : \schmC \to \catSet$ . Finally, denote by $\schmC^{op}$ the opposite category of $\schmC$ ; we will not use the term of contravariant functors $\schmC \to \schmD$ , but treat them as functors $\schmC^{op} \to \schmD$ .

Let us begin withe the definition of representable functors.

Definition.

We say that a functor $F : \schmC^{op} \to \catSet$ is represented by an object $X \in \schmC$ if there exists a natural equivalence $\eta : F \to \Hom_\schmC(-, X)$ .

If $F : \schmC^{op} \to \schmD$ is a functor, $X \in \schmC$ and $\schmD$ is concrete, we say that is represented by if $Forget \circ F : \schmC^{op} \to \catSet$ is represented by . Finally, we say that is representable if an object $X \in \schmC$ exists such that is represented by .
We say that a functor $F : \schmC \to \catSet$ is represented by an object $X \in \schmC$ if there exists a natural equivalence $\eta : F \to \Hom_\schmC(X, -)$ .

If $F : \schmC \to \schmD$ is a functor, $X \in \schmC$ and $\schmD$ is concrete, we say that is represented by if $Forget \circ F : \schmC \to \catSet$ is represented by . Finally, we say that is representable if an object $X \in \schmC$ exists such that is represented by .

This sounds rather abstract. Let us describe $\Hom_\schmC(-, X) : \schmC^{op} \to \catSet$ and $\Hom_\schmC(X, -) : \schmC \to \catSet$ in a little more in detail. To an object $A \in \schmC$ , they assign the set $\Hom_\schmC(A, X)$ of all morphisms from to , respectively the set $\Hom_\schmC(X, A)$ of all morphisms from to . And to a morphism $\varphi : A \to A'$ they assign the map

$\Hom_\schmC(\varphi, X) : \Hom_\schmC(A', X) \to \Hom_\schmC(A, X), \quad f \mapsto f \circ \varphi$

respectively

$\Hom_\schmC(X, \varphi) : \Hom_\schmC(X, A) \to \Hom_\schmC(X, A'), \quad f \mapsto \varphi \circ f.$

We want to make both concepts more clear with two important examples.

Example.

Let us consider the category $\schmC = \catRing(K)$ of -algebras, where is a fixed ring (or field, if you feel more comfortable). Let be the ring itself, and let $A = K[x_1, \dots, x_n]$ and $A' = K[x_1, \dots, x_n]/(f_1, \dots, f_m)$ , where $f_1, \dots, f_m \in K[x_1, \dots, x_n]$ are polynomials. Now every -homomorphism $\varphi : A \to K$ is a substitution homomorphism, i.e. there exists a uniquely determined $a = (a_1, \dots, a_n) \in K^n$ with $\varphi(f) = f(a_1, \dots, a_n)$ . Conversely, for any tuple $a = (a_1, \dots, a_n) \in K^n$ , there exists a homomorphism $\varphi_a : A \to K$ , $f \mapsto f(a)$ . Hence, we can identify $\Hom_\schmC(A, K)$ with .

Now the homomorphisms $A' = A/I \to K$ correspond to the homomorphisms $A \to K$ whose kernel contain . Let $\varphi_a : A \to K$ , $f \mapsto f(a)$ be a homomorphism. Then

$I \subseteq \ker \varphi_a \Leftrightarrow \forall i : f_i(a) = 0 \Leftrightarrow a \in V(f_1, \dots, f_m),$

where

$V := V(f_1, \dots, f_m) := \{ a \in K^n \mid f_1(a) = \dots = f_m(a) = 0 \}$

is the variety defined by the polynonials $f_1, \dots, f_m$ . Hence, we can identify $\Hom_\schmC(A', K)$ with . Now consider the projection $\pi : A \to A'$ , $f \mapsto f + I$ . If $\varphi \in \Hom_\schmC(A', K)$ corresponds to the point $a \in V$ , then $\Hom_\schmC(\pi, K)(\varphi) = \varphi \circ \pi : A \to K$ also corresponds to , but this time $a \in K^n$ ! Hence, $\Hom_\schmC(\pi, K)(\varphi)$ is the inclusion map $V \injto K^n$ .

Now, if one replaces $\Hom_\schmC(-, K)$ by $\Hom_\schmC(-, S)$ , where is a -algebra, we can identify $\Hom_\schmC(A, S)$ with and $\Hom_\schmC(A', S)$ with

$V_S := V_S(f_1, \dots, f_m) := \{ a \in S^n \mid f_1(a) = \dots = f_m(a) = 0 \},$

and again $\Hom_\schmC(\pi, S)$ is the inclusion map $V_S \injto S^n$ . Note that this is not just a toy example, but a very fundamental concept used with affine schemes. In fact, there, one fixes a -algebra , say our $A' = K[x_1, \dots, x_n]/(f_1, \dots, f_m)$ , and looks at the functor $\Hom_\schmC(A', -) : \schmC \to \catSet$ ; this functor is called the functor of points as it assigns to every -algebra (indirectly) the solutions $V_S(f_1, \dots, f_m)$ of the polynomial equations $f_1(a) = \dots = f_m(a) = 0$ with $a \in S^n$ . In fact, if we consider the functor

$V_\bullet : \schmC \to \catSet, \quad S \mapsto V_S(f_1, \dots, f_m)$

with $V_\phi : V_S \to V_{S'}$ , $(a_1, \dots, a_n) \mapsto (\phi(a_1), \dots, \phi(a_n))$ for $\phi : S \to S'$ , then we obtain a natural transformation

$\eta : V_\bullet \to \Hom_\schmC(A', \bullet)$

which assigns to $S \in \schmC$ the map

$V_S \to \Hom_\schmC(A', S), \quad a \mapsto \begin{cases} A' \to S \atop f \mapsto f(a). \end{cases}$

Since by the above, this is a bijection, it turns out that $\eta$ is in fact a natural equivalence; hence, represents the functor $V_\bullet$ .

Example.

Another example also comes from algebraic geometry, namely elliptic curves. If is an elliptic curve defined over a field , i.e. $a, b \in K$ , then for every field extension we can define

$E(L) := \{ (x, y) \mid y^2 = x^3 + a x + b \} \cup \{ \infty \}.$

For every -homomorphism $\phi : L \to L'$ , we obtain a map $E(\phi) : E(L) \to E(L')$ , $(x, y) \mapsto (\phi(x), \phi(y))$ , $\infty \mapsto \infty$ . Then $E : \catFld(K) \to \catAb$ is a functor into the category of abelian groups!

Now one can ask whether the functor is representable, i.e. whether there exists a field extension $L \in \catFld(K)$ of such that $\Hom_{\catFld(K)}(L', L)$ can be identified in a natural way with for every field extension – that is exactly what it means for being represented by . Unfortunately, it turns out not to be possible.

For two categories $\schmC$ , $\schmD$ , we can consider the category $\Hom(\schmC^{op}, \schmD)$ , whose objects are functors $\schmC^{op} \to \schmD$ and whose morphisms are natural transformations of such functors.

If we fix a category $\schmC$ and an object , we obtain the functor $h_X := \Hom_\schmC(-, X)$ , which is an element of $\Hom(\schmC^{op}, \catSet)$ . If we have another object together with a morphism $f : X \to Y$ , we obtain a natural transformation $h_f : h_X \to h_Y$ by assigning $Z \in \schmC$ the morphism $\Hom_\schmC(Z, X) \to \Hom_\schmC(Z, Y)$ , $g \mapsto f \circ g$ . Hence, can be seen as a functor from $\schmC$ to $\Hom(\schmC^{op}, \catSet) =: \hat{\schmC}$ . This functor is also called the (contravariant) Yoneda embedding.

Theorem (Yoneda's Lemma).

Let $\schmC$ be a category and $X \in \schmC$ , $H \in \hat{\schmC}$ . Then there is a natural bijection

$H(X) \to \Hom_{\hat{\schmC}}(h_X, H)$

with the following property:

If $u \in H(X)$ , then is mapped onto the natural transformation $h_X \to H$ , which satisfies that $g \in h_X(A)$ is mapped onto $H(g)(u) \in H(A)$ .

In other terms:

$\eta :{} & \bullet_2(\bullet_1) \to \Hom_{\hat{\schmC}}(h(\bullet_1), \bullet_2), \\ & (X, H) \mapsto \begin{cases} H(X) \to \Hom_{\hat{\schmC}}(h_X, H) \\ u \mapsto \begin{cases} h_X \to H \\ A \mapsto \begin{cases} h_X(A) \to H(A) \\ g \mapsto H(g)(u). \end{cases} \end{cases} \end{cases}$

is a natural equivalence.

This looks rather complicated. A simpler corollary is:

Corollary.

The Yoneda embedding $h : \schmC \to \hat{\schmC}$ is fully faithful, i.e. for every $X, Y \in \schmC$ , the map $h : \Hom_\schmC(X, Y) \to \Hom_{\hat{\schmC}}(h_X, h_Y)$ is bijective. In fact, is a full embedding.

Before we continue with the proof of Yoneda's lemma and the corollary, let us first take apart the notation from Yoneda's lemma. If $H \in \hat{\schmC}$ , it means that is a functor $\schmC^{op} \to \catSet$ . The class (which, by Yoneda's lemma, is a set) $\Hom_{\hat{\schmC}}(h_X, H)$ is the class of natural transformations $h_X \to X$ . Now, Yoneda's lemma basically says that there is a natural bijection between the set and the set of natural transformations $h_X \to H$ .

The property says that the bijection has to look as follows: to an element $u \in H(X)$ , we want to assign a natural transformation $\eta_u : h_X \to H$ which is defined as follows: for an object $A \in \schmC$ , we want that $\eta_u(A) : h_X(A) \to H(A)$ is defined by $g \mapsto H(g)(u)$ , i.e. we want to map $g \in \Hom_\schmC(A, X)$ to $H(g) : H(X) \to H(A)$ evaluated at , which was an element of .

Still sounds really complicated, doesn't it? Well, lets start with the proof, which includes some fancy diagrams.

Proof (of Yoneda's Lemma).

First, let $u \in H(X)$ . For $A \in \schmC$ , define $\eta_u(A) : h_X(A) \to H(A)$ by $g \mapsto H(g)(u)$ . Now let $\varphi : A \to A'$ be a morphism; then we have the following diagram:

$\xymatrix{ g \ar@{|->}[rrr] \ar@{|->}[dddd] & & & g \circ \varphi \ar@{|->}[ddd] \\ & \Hom_\schmC(A, X) \ar@{=}[d] & \Hom_\schmC(A', X) \ar@{=}[d] & \\ & h_X(A) \ar[r]^{h_\varphi} \ar[d]_{\eta_u(A)} & h_X(A') \ar[d]^{\eta_u(A')} & \\ & H(A) \ar[r]_{H(\varphi)} & H(A') & H(g \circ \varphi)(u) \\ H(g)(u) \ar@{|->}[rr] & & H(\varphi)( H(g)(u) ) & }$

But since is a contravariant functor, $H(g \circ \varphi) = H(\varphi) \circ H(g)$ ; therefore, $H(\varphi)(H(g)(u)) = H(g \circ \varphi)(u)$ . This shows that $\eta_u$ is a natural transformation, i.e. $\eta_u \in \hat{\schmC}$ .

Next, we have to show that

$\eta :{} & \bullet_2(\bullet_1) \to \Hom_{\hat{\schmC}}(h(\bullet_1), \bullet_2), \\ & (X, H) \mapsto \begin{cases} H(X) \to \Hom_{\hat{\schmC}}(h_X, H) \\ u \mapsto \eta_u \end{cases}$

is a natural equivalence between functors

$\bullet_2(\bullet_1) : \schmC \times \hat{\schmC} \to \catSet, \quad (X, H) \mapsto H(X)$

and

$\Hom_{\hat{\schmC}}(h(\bullet_1), \bullet_2) : \schmC \times \hat{\schmC} \to \catSet, \quad (X, H) \mapsto \Hom_{\hat{\schmC}}(h_X, H).$

For that, let $X, X' \in \schmC$ , $H, H' \in \hat{\schmC}$ and $\varphi : X \to X'$ a morphism and $\psi : H \to H'$ a natural transformation, and consider the following two diagrams:

$\xymatrix{ u \ar@{|-->}[ddd] \ar@{|-->}[rrr] & & & H(\varphi)(u) \ar@{|-->}[dd] \\ & H(X) \ar[r]^{H(\varphi)} \ar[d]_{\eta(X, H)} & H(X') \ar[d]^{\eta(X', H)} & \\ & \Hom_{\hat{\schmC}}(h_X, H) \ar[r]_{\Hom_{\hat{\schmC}}(h_\varphi, H)} & \Hom_{\hat{\schmC}}(h_{X'}, H) & \eta_{H(\varphi)(u)} \\ \eta_u \ar@{|-->}[rr] & & \eta_u \circ h_\varphi & }$

Now $\eta_u \circ h_\varphi$ and $\eta_{H(\varphi)(u)}$ are both natural transformations of functors $\schmC^{op} \to \catSet$ . Hence, let $A \in \schmC$ and $g \in h_X(A) = \Hom_\schmC(A, X)$ . Then

$\eta_{H(\varphi)(u)}(A)(g) = H(g)( H(\varphi)(u) ) = H(\varphi \circ g)(u)$

as is contravariant, and

$(\eta_u \circ h_\varphi)(A)(g) = \eta_u(A) \circ h_\varphi(A)(g) = \eta_u(A) \circ \varphi \circ g = H(\varphi \circ g)(u).$

Now consider the following diagram:

$\xymatrix{ u \ar@{|-->}[ddd] \ar@{|-->}[rrr] & & & \psi(X)(u) \ar@{|-->}[dd] \\ & H(X) \ar[r]^{\psi(X)} \ar[d]_{\eta(X, H)} & H'(A) \ar[d]^{\eta(X, H')} & \\ & \Hom_{\hat{\schmC}}(h_X, H) \ar[r]_{\Hom_{\hat{\schmC}}(h_X, \psi)} & \Hom_{\hat{\schmC}}(h_X, H') & \eta_{\psi(X)(u)} \\ \eta_u \ar@{|-->}[rr] & & \psi \circ \eta_u & }$

Now $\psi \circ \eta_u$ and $\eta_{\psi(X)(u)}$ are both natural transformations of functors $\schmC^{op} \to \catSet$ . Hence, let $A \in \schmC$ and $g \in h_X(A) = \Hom_\schmC(A, X)$ . Then

$\eta_{\psi(X)(u)}(g) = H'(g)( \psi(X)(u) ) = H'(g) \circ \psi(X)(u)$

and

$(\psi \circ \eta_u)(A)(g) = \psi(A) \circ \eta_u(A)(g) = \psi(A) \circ H(g)(u).$

Hence, the condition that these two elements are the same is equivalent to the fact that the diagram

$\xymatrix{ H(X) \ar[d]_{\psi(X)} \ar[r]^{H(g)} & H(A) \ar[d]^{\psi(A)} \\ H'(X) \ar[r]_{H'(g)} & H'(A) }$

commutes; but that follows from the fact that $\psi$ is a natural transformation $H \to H'$ . Hence, we have shown that $\eta$ is a natural transformation.

It is left to show that $\eta$ is in fact an equivalence, i.e. that for $X \in \schmC$ , $H \in \hat{\schmC}$ , the map

$\eta(X, H) : H(X) \to \Hom_{\hat{\schmC}}(h_X, H), \quad u \mapsto \eta_u$

is bijective. First, we show that it is injective; for that, note that for $u \in H(X)$ we have $\eta_u(X)(\id_X) = H(\id_X)(u) = \id_{H(X)}(u) = u$ , whence $\eta_u$ uniquely determines . To show that $\eta(X, H)$ is surjective, let $\eta' \in \Hom_{\hat{\schmC}}(h_X, H)$ . Set $u := \eta'(X)(\id_X)$ ; if $A \in \schmC$ is an object and $g \in h_X(A) = \Hom_\schmC(A, X)$ , consider the following commutative diagram:

$\xymatrix{ \id_X \ar@{|->}[rrr] \ar@{|->}[ddd] & & & g \ar@{|->}[dd] \\ & \Hom_\schmC(X, X) \ar[r]^{\Hom_\schmC(g, X)} \ar[d]_{\eta'(X)} & \Hom_\schmC(A, X) \ar[d]^{\eta'(A)} & \\ & H(X) \ar[r]_{H(g)} & H(A) & \eta'(A)(g) \\ u \ar@{|->}[rr] & & H(g)(u) & }$

Hence, $\eta'(A)(g) = H(g)(u) = \eta_u(A)(g)$ . Since and were arbitrary, we get $\eta' = \eta_u$ .

Wow, looks like a huge collection of abstract nonsense, eh? Well, the good news is that the worst part is done. Now, let us prove the corollary.

Proof (of the Corollary).

Let $X, Y \in \schmC$ , and let $H := h_Y \in \hat{\schmC}$ ; then $H(X) = \Hom_\schmC(X, Y)$ , and by Yoneda's Lemma, the map

$\eta(X, H) : \Hom_\schmC(X, Y) = H(X) \to \Hom_{\hat{\schmC}}(h_X, H) = \Hom_{\hat{\schmC}}(h_X, h_Y)$

is bijective. We have to check that $\eta(X, H)$ equals the map $h : \Hom_\schmC(X, Y) \to \Hom_{\hat{\schmC}}(h_X, h_Y)$ ; for that, let $u : X \to Y$ be a morphism, $A \in \schmC$ and $g \in h_X(A)$ . Then

$h(u)(A)(g) ={} & h_u(A)(g) = \Hom_\schmC(A, u)(g) = u \circ g \\ {}={} & \Hom_\schmC(g, Y)(u) = h_Y(g)(u) = \eta(X, H)(u)(A)(g).$

Finally, we have to show that is injective on objects. For $X \in \schmC$ , we have $h_X(X) = \Hom_\schmC(X, X)$ . Since morphism sets for distinct objects are assumed to be disjunct, it follows that we can get back from .

Well. After all this abstract nonsense, let us do some more concrete abstract nonsense. First, a lengthy definition which will turn out to be quite cool.

Definition.

Let $\schmC$ be a category with a final object in which finite products $G \times G$ and $G \times G \times G$ exist for all objects $G \in \schmC$ .

An object $G \in \schmC$ together with morphisms $m : G \times G \to G$ , $i : G \to G$ and $e : S \to G$ is called a group object if the following diagrams commute:

$\xymatrix@C+1cm{ G \times G \times G \ar[r]^{m \times \id_G} \ar[d]_{\qquad \id_G \times m} & G \times G \ar[d]^m \\ G \times G \ar[r]_m & G }$

$\xymatrix@C+0.5cm{ G \ar[r]^{(\id_G, i) \qquad} \ar[d] & G \times G \ar[d]^m \\ S \ar[r]_e & G } \qquad \xymatrix@C+0.5cm{ G \ar[r]^{(i, \id_G) \qquad} \ar[d] & G \times G \ar[d]^m \\ S \ar[r]_e & G }$

$\xymatrix@C-0.5cm{ G \times S \ar[rr]^{\id_G \times e} \ar[dr]_\cong & & G \times G \ar[dl]^m \\ & G & } \qquad \xymatrix@C-0.5cm{ S \times G \ar[rr]^{e \times \id_G} \ar[dr]_\cong & & G \times G \ar[dl]^m \\ & G & }$
We say that a group object is commutative if $m \circ w = m$ , where $w : G \times G \to G \times G$ switches its operands.
Let and be group objects. A morphism $\varphi : G \to H$ is called a homomorphism of group objects $\varphi : (G, m_G, i_G, e_G) \to (H, m_H, i_H, e_H)$ if $m_H \circ (\varphi \times \varphi) = \varphi \circ m_G$ , $i_H \circ \varphi = \varphi \circ i_G$ and $e_H = \varphi \circ e_G$ .

Let us first give some examples:

Example.

Let $\schmC = \catSet$ ; then $\{ 1 \}$ is a final object in $\schmC$ . The group objects in $\schmC$ are exactly the groups. A group object is commutative if the corresponding group is commutative, and homomorphisms of group objects are nothing else than group homomorphisms.

Example.

If $\schmC$ is the category of topological spaces, with continuous maps as morphisms, then the group objects in $\schmC$ are exactly the topological groups. Homomorphisms of group objects are again nothing else than continuous group homomorphisms.

Example.

If $\schmC = \catGrp$ is the category of all groups, the group objects in $\schmC$ are a bit more interesting. A final object is given by $S = (\{ 1 \}, \cdot)$ , a group of one element (in fact, is an initial object as well).

First, let $(A, \cdot)$ be an abelian group. Then $m : A \times A \to A$ , $(x, x') \mapsto x x'$ is a group homomorphism. So is $i : A \to A$ , $x \mapsto x^{-1}$ , and the map $e : S \to A$ , $1 \mapsto 1_A$ is a group homomorphism as well. Hence, every abelian group $(A, \cdot)$ gives rise to a commutative group object $((A, \cdot), m, i, e)$ .

Now, let $((G, \cdot), m, i, e)$ be a group object. As is a group homomorphism, we have . Next, since $m : G \times G \to G$ is a group homomorphism, we get

i.e. $m = \cdot$ . Now, $i : G \to G$ must be a group homomorphism as well, and the group object axioms force $i(g) = g^{-1}$ for all $g \in G$ ; hence, for all $g, h \in G$ ,

$g h = i(g^{-1}) i(h^{-1}) = i(g^{-1} h^{-1}) = (g^{-1} h^{-1})^{-1} = h g.$

Therefore, $((G, \cdot), m, i, e)$ forces $(G, \cdot)$ to be abelian, and the argument shows that , and are uniquely determined by $(G, \cdot)$ .

Therefore, the group objects in $\catGrp$ are exactly the abelian groups, and all of them are commutative. The group object homomorphisms between two abelian groups (with their only possible group object structure) are exactly the morphisms in $\catGrp$ between the two objects.

Now, one can do other stuff with group objects . Namely, given any other object $X \in \schmC$ , we can turn $\Hom_\schmC(X, G)$ into a group:

Proposition.

Let be a group object in $\schmC$ . For $X \in \schmC$ , set $H := \Hom_\schmC(X, G)$ and define

$m_H :{} & H \times H \to H, \quad (f, g) \mapsto m \circ (f \times g), \\ i_H :{} & H \to H, \quad f \mapsto i \circ f$

and set $e_H := e \circ \pi_X$ , where $\pi_X : X \to S$ is the unique morphism to the final object . Then is a group whose inverses are given by and whose neutral element is . If is commutative, then is abelian.

Proof.

First, we show that is associative. For that, let $f, g, h \in \Hom_\schmC(X, G)$ . Then

$m_H(m_H(f, g), h) ={} & m_H(m \circ (f \times g), h) \\ {}={} & m \circ ((m \circ (f \times g)) \times h) \\ \text{and} \quad m_H(f, m_H(g, h)) ={} & m_H(f, m \circ (g \times h)) \\ {}={} & m \circ (f \times (m \circ (g \times h)).$

We can rewrite this to

$m_H(m_H(f, g), h) ={} & m \circ (m \times \id_G) \circ (f \times g \times h) \ \text{and} \quad m_H(f, m_H(g, h)) ={} & m \circ (\id_G \times m) \circ (f \times g \times h),$

but from the definition of a group object, we know $m \circ (m \times \id_G) = m \circ (\id_G \times m)$ . Therefore, both expressions are the same.

Next, let us show that is a neutral element. For that, let $f \in \Hom_\schmC(X, G)$ . Then

$m_H(f, e_H) ={} & m \circ (f \times e_H) = m \circ ((\id_G \circ f) \times (e \circ \pi_X)) \\ {}={} & m \circ (\id_G \times e) \circ (f \times \pi_X)$

and, analogous, $m_H(e_H, f) = m \circ (e \times \id_G) \circ (\pi_X \times f)$ . But from the commutative diagrams of the definition of a group object, it turns out that both are the same .

Now let us show that is the inverse map. For that, let $f \in \Hom_\schmC(X, G)$ . Then

$m_H(f, i_H(f)) = m \circ (f \times (i \circ f)) = m \circ (\id_G \times i) \circ f$

and $m_H(i_H(f), f) = m \circ (i \times \id_G) \circ f$ . Now the definition of a group object gives $m \circ (\id_G \times i) = m \circ (i \times \id_G) = e \circ \pi_G$ , whence

$m_H(f, i_H(f)) = m_H(i_H(f), f) = e \circ \pi_G \circ f = e \circ \pi_X = e_H.$

Finally, assume that is commutative. Let $f, g \in \Hom_\schmC(X, G)$ ; then

$m_H(f, g) ={} & m \circ (f \times g) = m \circ w \circ (f, g) \\ {}={} & m \circ (g, f) = m_H(g, f);$

hence, is abelian.

For $\schmC = \catSet$ and $(G, \cdot)$ being a group (identified with the associated group object in $\schmC$ ), we get that the group $\Hom_\catSet(X, G)$ defined in the proposition is exactly the set $G^X = \{ (g_i)_{x \in X} \mid g_i \in G \}$ , where for elements $(g_x)_{x\in X}, (h_x)_{x \in X} \in G^X$ we define $(g_x)_{x \in X} (h_x)_{x \in X} := (g_x h_x)_{x \in X}$ ; then $(g_x)_{x \in X}^{-1} = (g_x^{-1})_{x \in X}$ and $1_{G^X} = (1_G)_{x \in X}$ .

In $\catSet$ , this seems to be not too interesting, but now assume that $\schmC$ is the category of topological spaces (with continuous maps). If $X = \{ p \}$ , then $\Hom_\schmC(X, G)$ gives exactly the group structure of , throwing away the additional information (i.e. that the group operations are continuous with respect to the topology on ). Now, if $X = \R$ with the usual topology, the elements of $\Hom_\schmC(X, G)$ are continuous paths in . Hence, in $\Hom_\schmC(X, G)$ , we can add continuous (parameterized) paths in (using pointwise addition)!

Now the interesting result is the following application of Yondea's lemma:

Theorem.

Let $\schmC$ be a category with finite products and a final object .

If is a group object, then the assignment

$\schmC \ni X \mapsto (\Hom_\schmC(X, G), (f, g) \mapsto m \circ (f \times g))$

is a functor $\schmC \to \catGrp$ . Moreover, is commutative if, and only if, the image of this functor lies in the subcategory $\catAb$ of $\catGrp$ .
Let and be two group objects in $\schmC$ , and let $F_G, F_H : \schmC \to \catGrp$ be the corresponding functors. The group object homomorphisms $\varphi : (G, m_G, i_G, e_G) \to (H, m_H, i_H, e_H)$ correspond one-to-one to natural transformations $F_G \to F_H$ .

In particular, two group object structures on an object give two naturally equivalent functors if, and only if, the group object structures are isomorphic.
Let $G \in \schmC$ . There is a one-to-one correspondence between the isomorphism classes of group object structures on and natural equivalence classes of functors $\schmC \to \catGrp$ which are represented by :

If is a group object structure, then

$X \mapsto (\Hom_\schmC(X, G), (f, g) \mapsto m \circ (f \times g))$

is the corresponding functor.

Before showing the theorem, note that the functor $h : \schmC \to \hat{\schmC}$ preserves products, i.e. if $X, Y \in \schmC$ and $X \times Y$ exists, then $h(X) \times h(Y)$ exists in $\hat{\schmC}$ and $h(X \times Y) \cong h(X) \times h(Y)$ in a natural way.

Proof.

First, we have to show that for morphisms $\varphi : X \to X'$ , the induced map $\Hom_\schmC(\varphi, G) : \Hom_\schmC(X, G) \to \Hom_\schmC(X', G)$ is a group homomorphism. For that, let $f, g \in \Hom_\schmC(X, G)$ ; then

$& \Hom_\schmC(\varphi, G)(m \circ (f \times g)) = m \circ (f \times g) \circ \varphi \\ {}={} & m \circ ((f \circ \varphi) \times (g \circ \varphi)) \ {}={} & m \circ (\Hom_\schmC(\varphi, G)(f) \times \Hom_\schmC(\varphi, G)(g)),$

what we had to show.

Finally, we are left to show that the fact that the image of the functor lies in $\catAb$ implies that is commutative, i.e. that $m \circ w = m$ . For that, consider the natural transformations $\hat{m} : h_G \times h_G \to h_G$ with $\hat{m}(X) = \cdot_{h_G(X)}$ and $\hat{w} : h_G \times h_G \to h_G \times h_G$ with $\hat{w}(X)(x, y) = (y, x)$ . (The fact that this is natural follows from the fact that we interpreted as a functor $\schmC \to \catGrp$ .) Then we have the commutative diagram

$\xymatrix{ h_G \times h_G \ar[rr]^{\hat{w}} \ar[dr]_{\hat{m}} & & h_G \times h_G \ar[dl]^{\hat{m}} \\ & h_G & }$

since for every $X \in \schmC$ , $(h_G(X), \cdot_{h_G(X)})$ is abelian. But now $h_G \times h_G = h_{G \times G}$ , whence this diagram is the image of the diagram

$\xymatrix{ G \times G \ar[rr]^w \ar[dr]_m & & G \times G \ar[dl]^m \\ & G & }$

under : obviously, $h(w) = \hat{w}$ and $h(m) = \hat{m}$ . But since is faithful, it follows from $\hat{m} \circ \hat{w} = \hat{m}$ that we also have $m \circ w = m$ , i.e. that is commutative.
By Yoneda's lemma, we get a bijection between the morphisms $G \to H$ and the natural transformations $Forget \circ F_G \to Forget \circ F_H$ . Hence, we have to sow that a morphism $\varphi : G \to H$ is a homomorphism of group objects if, and only if, the corresponding natural transformation $h_\varphi$ is actually a natural transformation $F_G \to F_H$ .

First, note that $\varphi$ is a homomorphisms of group objects and if, and only if, $m_H \circ (\varphi \times \varphi) = \varphi \circ m_G$ : in case this condition holds, one can obtain $i_H \circ \varphi = \varphi \circ i_G$ and $e_H = \varphi \circ e_G$ since are uniquely determined by and are uniquely determined by . (Just consider the same statement for groups: given the group operation, there is exactly one neutral element with respect to that operation and the inverses are unique as well.)

Now, since is faithful, the diagram

$\xymatrix{ G \times G \ar[r]^{m_G} \ar[d]_{\varphi \times \varphi} & G \ar[d]^\varphi \\ H \times H \ar[r]_{m_H} & H }$

commutes if, and only if, the diagram

$\xymatrix{ h_G \times h_G \ar[r]^{h_{m_G}} \ar[d]_{h_\varphi \times h_\varphi} & h_G \ar[d]^{h_\varphi} \\ h_H \times h_H \ar[r]_{h_{m_H}} & h_H }$

commutes. That the first diagram commutes is equivalent to the fact that $\varphi$ is a homomorphism of group objects. That the second diagram commuts is equivalent to the fact that $h_\varphi(X)$ is a group homomorphism for every $X \in \schmC$ , i.e. that $h_\varphi$ is a natural transformation for functors $\schmC \to \catGrp$ .
Part 1. shows that every group object structure on induces a functor $\schmC \to \catGrp$ . Part 2. shows the statement that the group structures are isomorphic if, and only if, the functors are naturally equivalent.

Let $F : \schmC \to \catGrp$ be a functor which is represented by , i.e. there exists a natural equivalence $\eta : \Hom_\schmC(-, G) \to Forget \circ F$ . For $X \in \schmC$ , define

$\hat{m}_X :{} & \Hom_\schmC(A, G) \times \Hom_\schmC(A, G) \to \Hom_\schmC(A, G) \\ & (f, g) \mapsto \eta(A)^{-1}(\eta(A)(f) \cdot_{F(A)} \eta(A)(f)).$

As $\eta$ is a natural transformation, it turns out that $\hat{m} : A \mapsto \hat{m}_X$ gives a natural transformation

$h_G \times h_G = \Hom_\schmC(-, G) \times \Hom_\schmC(-, G) \to \Hom_\schmC(-, G) = h_G.$

Since $h_G \times h_G \cong h_{G \times G}$ in a natural way, we have that $\hat{m}$ gives a natural transformation $h_{G \times G} \to h_G$ . By the corollary on Yoneda's lemma, this natural transformation corresponds to a morphism $m : G \times G \to G$ .

Let us show that $m : G \times G \to G$ satisfies the associativity diagram, i.e. we have that

$\xymatrix@C+1cm{ G \times G \times G \ar[r]^{m \times \id_G} \ar[d]_{\qquad \id_G \times m} & G \times G \ar[d]^m \\ G \times G \ar[r]_m & G }$

commutes.

Consider the natural transformations

$\hat{m} \circ \id :{} & (h_G \times h_G) \to h_G \to h_G \times h_G \\ \text{and} \quad \id \times \hat{m} :{} & h_G \times (h_G \times h_G) \to h_G \times h_G.$

From the definition of $\hat{m}$ it follows that the following diagram commutes:

$\xymatrix{ h_G \times h_G \times h_G \ar[r]^{\hat{m} \times \id} \ar[d]_{\id \times \hat{m}} & h_G \times h_G \ar[d]^{\hat{m}} \\ h_G \times h_G \ar[r]_{\hat{m}} & h_G }$

(Simply plug in an object $X \in \schmC$ , and then use the definition of $\hat{m}$ and the fact that $(F(X), \cdot_{F(X)})$ is a group.) Now, since $h_G \times h_G = h_{G \times G}$ and $h_G \times h_G \times h_G = h_{G \times G \times G}$ , one can see that this diagram is the image of the previous diagram under . But since is faithful, the previous diagram also has to commute.

Now, one can obtain $i : G \to G$ and $e : S \to G$ in the same manner and show that they satisfy the conditions they have to such that is a group object.

In a nutshell, this result says that representable functors $\schmC \to \catGrp$ are the same as group objects in $\schmC$ , and that representable functors $\schmC \to \catAb$ are the same as commutative group objects in $\schmC$ . This is somewhat surprising, as one expects that constructing a group object structure on an object in $\schmC$ is hard, while coming up with a (representable) functor $\schmC \to \catGrp$ (or $\catAb$ ) sounds easier. In fact, constructing a functor $\schmC \to \catGrp$ is usually easier, but showing that the constructed functor is representable is hard. (Just consider the construction of Picard schemes or Hilbert schemes.)

If you are interested in literature, see, for example, the book “Néron Models” by S. Bosch, W. Lütkebohmert and M. Raynaud (Ergebnisse der Mathematik und ihrer Grenzgebiete no. 21, Springer, 1990), and the book “Commutative Group Schemes” by F. Oort (Lecture Notes in Mathematics no. 15, Springer, 1966).

Felix' Math Place

Focussed on, but not limited to Computational Number Theory

Fun With Representable Functors, or Why I Like Yondea's Lemma.

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Felix' Math Place

Focussed on, but not limited to Computational Number Theory

Fun With Representable Functors, or Why I Like Yondea's Lemma.

Comments.

About Me.

About This Blog.

Overview Pages.

Categories.

Recent Posts.

Archives.

Tags.

Impressum.

Data Protection.