Let $F:\mathscr{A} \to \mathscr{C}$ and $G:\mathscr{C} \to \mathscr{A}$ be an adjoint pair of functors.

I am trying to show $G$ preserves products and $F$ preserves coproducts.

So to start we have a bijection

$\tau_{AC}:\operatorname{Hom}(FA,C) \to \operatorname{Hom}(A,GC)$

as well as the normal naturality diagram for both variables.

It suffices to prove only one pair of the question - the other will follow form dualising the proof (or I am guessing working in the opposite categories or some other nice trick).

I decided to try show that $G$ preserves products. So we have the usual diagram for the product: i.e. given $C_1,C_2 \in \mathscr{C}$ there exists morphisms $g_1:C_1 \times C_2 \to C_1$ and $g_2:C_1 \times C_2 \to C_2$ such that for any other map $X \to C_i,i=1,2$ there is a unique map $\theta:X \to C_1 \times C_2$.

Now I can immediately apply the functor $G$ to this. This doesn't quite give what I want. In particular it gives $G(C_1 \times C_2)$ in the usual universal diagram. To solve the problem we need to replace that with $G(C_1) \times G(C_2)$.

I've played around for quite awhile to no real avail. Obviously I need to use naturality. So given $g:C \to C'$ in $\mathscr{C}$ we have that

$(Gg)_* \tau_{AC} = \tau_{AC'} g_*$.

where $$g_*:\operatorname{Hom}(FA,C) \to \operatorname{Hom}(FA,C')$$ and $$(Gg)_*: \operatorname{Hom}(A,GC) \to \operatorname{Hom}(A,GC').$$

I was thinking the way to go would be to use the maps $g_i: C_1 \times C_2 \to C_i$ and play around with the natural transformation diagram to get something nice, but I keep getting nowhere.

Any hints?

(I would add that I am aware there is a more general result that $F$ preserves colimits and $G$ preserves limits, but I am not familiar enough with them to work with them just yet.)


Do you know that $\hom(X,-)$ preserves products, don't you? Then, you can use the chain of isomorphism

$\hom(A,G(C_1\times C_2))\cong\hom(F(A),C_1\times C_2)\cong$


$\cong\hom(A,G(C_1))\times \hom(A,G(C_2))\cong \hom(A,G(C_1)\times G(C_2))$

Now if $\hom(A,X)\cong \hom(A,Y)$ for any $A$, then $X\cong Y$.

If not, I think proving that $\hom(-,-)$ preserves products and turn coproducts into products might be more useful and easy.

  • $\begingroup$ I didn't make precise the last sentence, and if you are a beginner there's no reason to throw the claim on your face without explain a little bit: what I mean is that $\hom(X,Y_1\times Y_2)\cong \hom(X,Y_1)\times \hom(X,Y_2)$ and $\hom(X_1\amalg X_2,Y)\cong \hom(X_1)\times \hom(X_2,Y)$. $\endgroup$
    – fosco
    Jun 23 '11 at 8:31
  • $\begingroup$ nice proof! $\endgroup$
    – Juan S
    Jun 23 '11 at 12:37

You were close, but you missed the point of universality, so let me do the argument you had in mind, instead of a different one (the route chosen by tetrapharmakon). I agree that tetrapharmakon's argument is the efficient way to go, but I think it's a worthwhile exercise to do it without implicitly appealing to Yoneda. More to the point: Explicitly following the maps through all the natural isomorphisms in tetrapharmakon's answer is quite painful (at least to me), so a direct argument involving only the definitions makes me feel more comfortable.

Let me start from scratch (because I'm old and rigid, I'm unable to work with $F$ and $G$ in adjunctions, so please bear with me and let me replace them by the more descriptive letters $L$ and $R$).

So, we're given an adjunction $L : \mathscr{A} \longleftrightarrow \mathscr{C} : R$.

We're given $C,D \in \mathscr{C}$ and a product diagram $C\;\xleftarrow{p_C}\; C \times D\; \xrightarrow{p_D}\; D$. Applying the right adjoint $R$ we get the diagram $R(C)\; \xleftarrow{R(p_D)}\; R(C \times D) \;\xrightarrow{R(p_D)}\;R(D)$.

We want to see that this diagram is a product of $R(C)$ and $R(D)$. The only thing we need to do is to check the universal property.

So let us be given morphisms $R(C)\; \xleftarrow{a_C} \; A \; \xrightarrow{a_D} \; R(D)$ and we want to show that there is a unique morphism $d:A \to R(C \times D)$ such that $R(p_C)d = a_C$ and $R(p_D)d = a_D$. By adjointness $$\operatorname{Hom}_{\mathscr{C}}(LA, C \times D) = \operatorname{Hom}_{\mathscr{A}}(A,R(C\times D)),$$ so we can translate the problem into a problem in $\mathscr{C}$ by applying $L$. Thus, we consider the diagram $LR(C) \; \xleftarrow{L(a_C)}\; L(A) \;\xrightarrow{L(a_D)}\;LR(D)$. This is not quite where we want to be, but remembering the triangular identities for counit $\varepsilon: LR \Rightarrow 1_\mathscr{C}$ and unit $\eta:1_{\mathcal{A}} \Rightarrow RL$ leads us to $$C \; \xleftarrow{\varepsilon_C L(a_C)}\; L(A)\;\xrightarrow{\varepsilon_D L(a_D)}\;D.$$ Now, applying the universal property of the product diagram we started with, we finally find a unique morphism $e:L(A) \to C \times D$ such that $p_Ce = \varepsilon_C L(a_C)$ and $p_De= \varepsilon_D L(a_D)$.

Now the composition $d$ of $A \; \xrightarrow{\eta_{A}}\; RL(A)\;\xrightarrow{R(e)} \; R(C \times D)$ is the morphism we're looking for. Indeed, since $\eta : 1_{\mathcal{A}} \Rightarrow RL$ is a natural transformation, we have the commutative diagram

$$\begin{array}{ccc} A & \xrightarrow{a_C} & R(C) \\ \downarrow{\scriptstyle \eta_A} & & \downarrow{\scriptstyle \eta_{RC}} \\ RL(A) & \xrightarrow{RL(a_C)} & RLR(C) \end{array}$$

and combining this with the triangular identity $R(\varepsilon_C)\eta_{RC} = 1_{RC}$ we get

$$R(p_C)d = R(p_C)R(e)\eta_A = R(p_Ce) \eta_A = R(\varepsilon_C L(a_C)) \eta_A = R(\varepsilon_C) RL(a_C) \eta_A = R(\varepsilon_C) \eta_{RC} a_C = a_C$$

and similarly $R(p_D)d = a_D$. I leave it to you to convince yourself of the uniqueness of $d$.

Finally, let me stress that exactly the same argument works with general limits instead of binary products:

Given a diagram $D: \mathscr{D} \to \mathscr{C}$ with limit $C = \varprojlim_{\mathscr{D}} D$ (constant diagram) and universal morphism $u: C \Rightarrow D$, the morphism $R(u): R(C) \Rightarrow R \circ D$ exhibits $R(C)$ as $\varprojlim_{\mathscr{D}} R\circ D$. The only thing that needs to be checked is given any diagram $A: \mathscr{D} \to \mathscr{A}$ with morphism $a: A \Rightarrow R \circ D$, the morphism $a$ factors uniquely as $a = R(u)d$ for a morphism $d: A \Rightarrow R(C)$. The morphism $d$ is obtained as $R(e) \eta A$, where $e: L A \Rightarrow C$ is obtained from the universality of $u$ applied to $\varepsilon LA$.

As usual in category theory, this is all rather tautological (but admittedly confusing at first sight).

  • $\begingroup$ "because I'm old and rigid, I'm unable to work with $F$ and $G$" funny; I can only work with $F$ and $G$ for an adjunction, because that's how I learned it from Mac Lane :) $\endgroup$ Jun 23 '11 at 18:59
  • $\begingroup$ @wildildildlife: Heh! De gustibus... Seriously, I'm always getting confused when reading Mac Lane. Writing an adjunction $\langle F,G, \phi \rangle: X \to A$. So okay $F$ for free, but why $X$ and $A$? and $\phi$ goes from which hom set to which, exactly? So I had to sit down and do it myself and I did it with $L$ and $R$, that's why. $\endgroup$
    – t.b.
    Jun 23 '11 at 19:08
  • $\begingroup$ @Theo: I think you can avoid appealing Yoneda using a similar universality-argument, but your proof is so clean and precise it's not worth to add anything. :) $\endgroup$
    – fosco
    Jun 23 '11 at 19:09
  • $\begingroup$ @tetrapharmakon: Yes, you're right indeed. Thanks, by the way! :) $\endgroup$
    – t.b.
    Jun 23 '11 at 19:15
  • $\begingroup$ @Theo, thank you for taking the time to write this all up! This was indeed the way I was thinking the proof would go. $\endgroup$
    – Juan S
    Jun 23 '11 at 22:10

Suppose we have an adjunction as in your question $F \dashv G $, and a product $A\times B$ with projections $\pi_1 : A\times B\rightarrow A$ and $\pi_1 : A\times B\rightarrow B$. If we want to show $G$ preserves the product $A\times B$ we need only show that $G(A\times B)$ with the projections maps $G(\pi_1) : G(A\times B)\rightarrow G(A)$ and $G(\pi_1) : G(A\times B)\rightarrow G(B)$ has the universal property that the product $G(A) \times G(B)$ must have (if it exists) and by uniqueness they are the same.

Given maps $c_1 : C\rightarrow G(A)$ and $ c_2 : C\rightarrow G(B)$ via the adjunction we obtain maps $\bar c_1 :F(C)\rightarrow A$ and $\bar c_2 :F(C)\rightarrow B$ which, by the universal property of $A\times B$, induce a unique map $\bar u:F(C)\to A\times B$ such that $\bar c_i = \pi_i \bar u$ (i = 1,2). Going back through the adjunction we get a map $u: C \to G(A \times B)$ such that $c_i = G(\pi_i) u$ (i = 1,2) by naturality of the adjunction, and it is unique among such maps as the adjunction is a bijection of hom-sets so $G(A\times B)$ has the universal property that the product $G(A)\times G(B)$ must have so $G(A\times B) = G(A)\times G(B)$ hence $G$ preserves any products which exists in $\mathcal C$. Note that this argument easily generalizes to show that $G$ preserves all limits which exist in $\mathcal C$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.