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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.)

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3 Answers 3

up vote 7 down vote accepted

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

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

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

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

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.

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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 $\hom(X_1\amalg X_2,Y)\cong \hom(X_1)\times \hom(X_2,Y)$. –  tetrapharmakon Jun 23 '11 at 8:31
    
nice proof! –  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).

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"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 :) –  wildildildlife Jun 23 '11 at 18:59
    
@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. –  t.b. Jun 23 '11 at 19:08
    
@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. :) –  tetrapharmakon Jun 23 '11 at 19:09
    
@tetrapharmakon: Yes, you're right indeed. Thanks, by the way! :) –  t.b. Jun 23 '11 at 19:15
    
@Theo, thank you for taking the time to write this all up! This was indeed the way I was thinking the proof would go. –  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) but going back through the adjunction we get a map $c: C \to G(A \times B)$ such that $c_i = G(\pi_i) u$ (i = 1,2) by naturality of the adgunction, 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 $\mathcal C$

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