I've been having troubles trying to make sense of the last part of Exercise 7.11.8 in Awodey's Category Theory book (p.182). The exercise asks us to

  1. Show that a functor category $\mathbf{D}^\mathbf{C}$ has binary products if $\mathbf{D}$ does (construct the product of two functors F and G "objectwise": $(F×G)(C) = F(C) × G(C)$).

And defining the product for natural transformations, and showing that the construction satisfies the UMP is not difficult. That is, there is a unique $u_c : ZC \to F(C) \times G (C)$, given the UMP of the product in $\mathbf{D}$:enter image description here

But the point I am confused at, is that we need to show that $h_C$ is a natural transformation, i.e. that the diagram

$\qquad \qquad\qquad\qquad \qquad$enter image description here

commutes. And I have no idea how to do that. In fact, I don't even know what (the definition of) $Zf: ZC \to ZD$ or $Ff \times Gf$ would be.

Awodey gives a solution at p.295. But I can't get what he is doing. He says:

$\qquad \qquad $ enter image description here

Without justifying it any further.

What I want to understand is this:

why is that $\pi_1^{FD \times GD}\circ Ff \times Gf \circ h_C$ equals $Ff \circ \pi_1^{FC \times GC} \circ h_C$? It makes sense to me that $\pi_1^{FD \times GD}\circ Ff \times Gf = Ff$. But where is this $\pi_1^{FC \times GC} $ coming from?


  • $\begingroup$ Note that it isn't true that $\pi_1^{FD \times GD}\circ F(f)\times G(f)=F(f)$ because the first morphism belongs to $\mathbf D[F(C),F(D)]$ while the second belongs to $\mathbf D[F(C)\times G(C),F(D)]$. $\endgroup$ – Giorgio Mossa Jan 6 '16 at 20:15

OK, let's go through this step-by-step using the notation you set up in your first diagram.

We have two functors $F$ and $G$ from $\mathbf{C}$ to $\mathbf{D}$. We define a functor $F\times G$ such that $(F\times G)(C) = F(C) \times G(C)$. It's worth noting how this functor operates on arrows: given $f\colon C\to D$, we get arrows $F(f)\colon F(C)\to F(D)$ and $G(f)\colon G(C)\to G(D)$, so we can define $(F\times G)(f)$ to be the unique arrow $F(C)\times G(C)\to F(D)\times G(D)$ (arrows into a product are determined by their compositions with the projections) such that $$p_1^{F(D)\times G(D)}\circ (F\times G)(f) = F(f)\circ p_1^{F(C)\times G(C)} \text{ and } p_2^{F(D)\times G(D)}\circ (F\times G)(f) = G(f)\circ p_2^{F(C)\times G(C)}.$$

These equations show that we have two natural transformations $p_1\colon (F\times G) \to F$ and $p_2\colon (F\times G) \to G$, where $(p_1)_C$ is the projection $F(C)\times G(C)\to F(C)$ and similarly for $p_2$. We'd like to show that this data makes $F\times G$ the product of $F$ and $D$ in the functor category.

So we take another functor $Z$ and given natural transformations $z_1\colon Z\to F$ and $z_2\colon Z\to G$. For each object $C$, we get an arrow $h_C\colon Z(C)\to F(C)\times G(C)$ which is the unique arrow such that $(p_1)_C\circ h_C = (z_1)_C$ and $(p_2)_C\circ h_C = (z_2)_C$. We need to show that the $h_C$ cohere to a natural transformation.

That is, for every map $f\colon C\to D$, we must show that $h_D\circ Z(f) = (F\times G)(f)\circ h_C$. These are maps $Z(C)\to F(D)\times G(D)$, so by the universal property of the product, it suffices to show $$(p_1)_D\circ h_D\circ Z(f) = (p_1)_D\circ (F\times G)(f)\circ h_C \text{ and } (p_2)_D\circ h_D\circ Z(f) = (p_2)_D\circ (F\times G)(f)\circ h_C.$$

Now the answer to your question. We'll just do the $p_1$ case, since the $p_2$ case is exactly the same.

$(p_1)_D\circ h_D\circ Z(f) = (z_1)_D \circ Z(f)$. This is by definition of $h_D$ ($(p_1)_D\circ h_D = (z_1)_D$).

$(z_1)_D \circ Z(f) = F(f)\circ (z_1)_C$. This is what it means for $z_1$ to be a natural transformation.

$F(f)\circ (z_1)_C = F(f) \circ (p_1)_C\circ h_C$. Again, since $(p_1)_C\circ h_C = (z_1)_C$.

$F(f) \circ (p_1)_C\circ h_C = (p_1)_D\circ (F\times G)(f) \circ h_C$. This is by naturality of $p_1$, or, equivalently, the equations above defining the action of $F\times G$ on arrows.

And we're done! Awodey chases through the chain of equalities in the opposite order and with different notation, but the steps are exactly the same.

You can visualize what's going on diagrammatically by stacking two copies of your triangular diagram on top of one other, one with $C$s and one with $D$s, and connecting the top diagram to the bottom one with arrows $Z(f)$, $F(f)$, $G(f)$, and $F(f)\times G(f)$. Then we need to check that the central vertical square commutes, and we do this using commutativity of the all the vertical squares in the "wings".


$Z$ is a given functor, hence it is already defined on objects and arrows; you're baffled by the fact that it is not given explicitly.

But you don't need that: it's good old abstract nonsense :)

Awodey's trick is simple, in hindsight: when you want to prove that two arrows to a product are equal, it's enough to prove that they become equal whenever composed with each of the projection (so, the same argument works even with arbitrarily large products, and with general limits).

What I mean is that the universal property of the product (/the limit) says that $(Ff\times Gf)\circ h_C = h_D\circ Zf$ (i.e. $h$ is natural) iff $p_{1,C}\circ (Ff\times Gf)\circ h_C = p_{1,C}\circ h_D\circ Zf$ and $p_{2,C}\circ (Ff\times Gf)\circ h_C = p_{2,C}\circ h_D\circ Zf$. Basically because there can be a unique arrow with this property, so the two are in fact one.

That's what you are supposed to prove, and that's what Awodey does. Bye!

  • $\begingroup$ It's important to keep the same notation all along the proof; if you are uncomfortable with Awodey's, choose your favourite one! Are you sure that $\alpha$ is never mentioned? $\endgroup$ – Fosco Loregian Dec 30 '15 at 12:10
  • $\begingroup$ I'm starting a bounty for this question because I still don't understand Awodey's calculations, i.e. why is that $\pi_1^{FD \times GD}\circ Ff \times Gf \circ h_C$ equals $Ff \circ \pi_1^{FC \times GC} \circ h_C$ etc? $\endgroup$ – StudentType Jan 5 '16 at 12:50

By definition $F(f)\times G(f)$ is the only morphism such that the diagram $$\require{AMScd} \begin{CD} F(C) \times G(C) @>\pi_1^{F(C)\times G(C)}>> F(C) \\ @V F(f)\times G(f)VV @V F(f)VV\\ F(D) \times G(D) @>\pi_1^{F(D) \times G(D)}>> F(D)\\ \end{CD} $$ and the diagram $$ \begin{CD} F(C) \times G(C) @>\pi_2^{F(C)\times G(C)}>> G(C) \\ @V F(f)\times G(f)VV @V G(f)VV\\ F(D) \times G(D) @>\pi_2^{F(D) \times G(D)}>> G(D)\\ \end{CD} $$ commute.

This implies that $\pi_1^{F(D)\times G(D)}\circ F(f) \times G(f) = F(f) \circ \pi_1^{F(C) \times D(C)}$ and that $\pi_2^{F(D) \times G(D)}\circ F(f) \times G(f)=G(f) \circ \pi_2^{F(C)\times G(C)}$.

The equality you're asking for follows by precomposing the two equalities above with $h_C$.


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