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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:

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


“for every group, G, F(G) 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 K by \catFld(K) and, for a ring R, the category of R-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.

  1. 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 F is represented by X if Forget \circ F : \schmC^{op} \to \catSet is represented by X. Finally, we say that F is representable if an object X \in \schmC exists such that F is represented by X.

  2. 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 F is represented by X if Forget \circ F : \schmC \to \catSet is represented by X. Finally, we say that F is representable if an object X \in \schmC exists such that F is represented by X.

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 A to X, respectively the set \Hom_\schmC(X, A) of all morphisms from X to A. 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


\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.


Let us consider the category \schmC = \catRing(K) of K-algebras, where K is a fixed ring (or field, if you feel more comfortable). Let X = K 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 K-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 K^n.

Now the homomorphisms A' = A/I \to K correspond to the homomorphisms A \to K whose kernel contain I. 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),


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 V. 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 a, 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 S is a K-algebra, we can identify \Hom_\schmC(A, S) with S^n 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 K-algebra A', 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 K-algebra S (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, A' represents the functor V_\bullet.


Another example also comes from algebraic geometry, namely elliptic curves. If E : y^2 = x^3 + a x + b is an elliptic curve defined over a field K, i.e. a, b \in K, then for every field extension L/K we can define

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

For every K-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 E is representable, i.e. whether there exists a field extension L \in \catFld(K) of K such that \Hom_{\catFld(K)}(L', L) can be identified in a natural way with E(L') for every field extension L'/K – that is exactly what it means for F being represented by L'. 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 X, we obtain the functor h_X := \Hom_\schmC(-, X), which is an element of \Hom(\schmC^{op}, \catSet). If we have another object Y 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, h 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 u 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:


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, h 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 H 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 H(X) 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 u, which was an element of H(X).

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 H 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)


\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 H 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)


(\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 u. 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 A and g 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 h 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 X back from h_X(X).

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.


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

  1. 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 & }
  2. We say that a group object (G, m, i, e) is commutative if m \circ w = m, where w : G \times G \to G \times G switches its operands.
  3. Let (G, m_G, i_G, e_G) and (H, m_H, i_H, e_H) 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:


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.


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.


If \schmC = \catGrp is the category of all groups, the group objects in \schmC are a bit more interesting. A final object S is given by S = (\{ 1 \}, \cdot), a group of one element (in fact, S 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 e is a group homomorphism, we have e(1) = 1_G. Next, since m : G \times G \to G is a group homomorphism, we get

m(g, h) = m(g, 1) m(1, h) = g h,

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 m, i and e 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 (G, m, e, i). Namely, given any other object X \in \schmC, we can turn \Hom_\schmC(X, G) into a group:


Let (G, m, e, i) 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 S. Then (H, m_H) is a group whose inverses are given by i_H and whose neutral element is e_H. If (G, m, e, i) is commutative, then (H, m_H) is abelian.


First, we show that m_H 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 e_H 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 f.

Now let us show that i_H 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 (G, m, i, e) 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, (H, m_H) 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 G, throwing away the additional information (i.e. that the group operations are continuous with respect to the topology on G). Now, if X = \R with the usual topology, the elements of \Hom_\schmC(X, G) are continuous paths in G. Hence, in \Hom_\schmC(X, G), we can add continuous (parameterized) paths in G (using pointwise addition)!

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


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

  1. If (G, m, i, e) 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, (G, m, i, e) is commutative if, and only if, the image of this functor lies in the subcategory \catAb of \catGrp.

  2. Let (G, m_G, i_G, e_G) and (H, m_H, i_H, e_H) 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 G give two naturally equivalent functors if, and only if, the group object structures are isomorphic.

  3. Let G \in \schmC. There is a one-to-one correspondence between the isomorphism classes of group object structures on G and natural equivalence classes of functors \schmC \to \catGrp which are represented by G:

    If (G, m, i, e) 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.

  1. 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 (G, m, i, e) 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 h_G 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 h: obviously, h(w) = \hat{w} and h(m) = \hat{m}. But since h is faithful, it follows from \hat{m} \circ \hat{w} = \hat{m} that we also have m \circ w = m, i.e. that (G, m, i, e) is commutative.

  2. 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 (G, m_G, i_G, e_G) and (H, m_H, i_H, e_H) 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 i_G, e_G are uniquely determined by m_G and i_H, e_H are uniquely determined by m_H. (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 h 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.

  3. Part 1. shows that every group object structure on G 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 G, 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 }


    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 h. But since h 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 (G, m, i, e) 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).


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