# Definition of $[\mathscr A^{\text{op}}, \textbf{Set}](H_A,X)\to [\mathscr A^{\text{op}}, \textbf{Set}](H_B,X)$ in the proof of the Yoneda lemma

Regarding the proof of the Yoneda lemma on p.98:

How does one get that the map $$[\mathscr A^{\text{op}}, \textbf{Set}](H_A,X)\to [\mathscr A^{\text{op}}, \textbf{Set}](H_B,X)$$ is defined by $$-\circ H_f$$ ? (The definition of $$H_f$$ is given on p.90.)

And how does one get that the map $$[\mathscr A^{\text{op}}, \textbf{Set}](H_A,X)\to [\mathscr A^{\text{op}}, \textbf{Set}](H_A,X')$$ is defined by $$\theta\circ -$$ ?

• That's just applying the definition of what a Hom-functor is on $[\mathscr{A}^{op},\mathbf{Set}]$, unless I'm missing something about your question. Commented Jul 14, 2019 at 7:31
• @MaliceVidrine The hom functor would be $[\mathscr A^{\text{op}}, \textbf{Set}](-,X)$, I think our case is more complicated than that -- I guess the functor is $[\mathscr A^{\text{op}}, \textbf{Set}](H_\bullet,X)$, which is the composition of $H_\bullet$ with $[\mathscr A^{\text{op}}, \textbf{Set}](-,X)$. But I'm still confused: the former functor has codomain $[\mathscr A^{\text{op}}, \textbf{Set}]$, whereas the domain of the latter functor is $[\mathscr A^{\text{op}}, \textbf{Set}]^{op}$, so they cannot be composed. Commented Jul 14, 2019 at 15:26

For any category $$\mathcal C$$, the hom functor acts on morphisms by pre- and post composition: if $$\alpha:A'\to A$$ and $$\beta:B\to B'$$ are morphisms in $$\mathcal C$$, then $$\mathcal C(\alpha,\beta):\mathcal C(A,B)\to \mathcal C(A', B')\\ \varphi \ \mapsto\ \beta\circ\varphi\circ\alpha$$ And in particular, it's usual to write $$\mathcal C(A,\beta)$$ for $$\mathcal C(1_A,\beta)$$ and $$\mathcal C(\alpha, B)$$ for $$\mathcal C(\alpha, 1_B)$$.

Apply it with $$\mathcal C =[\mathscr A^{op}, \mathbf{Set}]$$, $$\ \alpha=H_f\$$ and $$\ \beta=\theta$$.

• I think this answers my second question, but not the first (see my comment above). Commented Jul 14, 2019 at 15:44
• No, it's the same: for $f:A\to B$ in $\mathscr A$ we get $H_f:H_A\to H_B$ (whose $X$ component is just $\mathscr A(X, f)$) and we apply the above with $\alpha =H_f$. Commented Jul 14, 2019 at 18:08
• I still feel such construction sweeps the details under the rug (probably because I don't have much experience with category theory and want to spell out every detail). I posted an answer which looks more plausible for me. Commented Jul 14, 2019 at 19:19

The single argument Hom-functors you're probably more familiar with are just what happens when you fix one of the arguments $$\mathcal{C}(-,-):\mathcal{C}^{op}\times\mathcal{C}\to\mathbf{Set}$$. As functors with that domain, you can absolutely use both the pre- and post-compositions this proof is talking about.

But I don't think you're quite clear on what the proof is trying to do. It's trying to establish a natural isomorphism between two functors $$\mathscr{A}^{op}\times[\mathscr{A}^{op},\mathbf{Set}]\to \mathbf{Set}$$. One of these functors is the evaluation functor, but the other is the composite $$\hom\circ (H_{-}\times id):\mathscr{A}^{op}\times[\mathscr{A}^{op},\mathbf{Set}]\to[\mathscr{A}^{op},\mathbf{Set}]^{op}\times[\mathscr{A}^{op},\mathbf{Set}]\to\mathbf{Set}$$ where that "$$\hom$$" is the two-argument Hom-functor. The first half of the stuff on p.98 is establishing naturality of this isomorphism in the first argument, and the second half is establish naturality in the second argument. But to talk about the behavior in both arguments of that big composite above is precisely to talk about instances of both pre- and post-composition.

First question. The context is that we want to prove that $$[\mathscr A^{\text{op}}, \textbf{Set}](H_A,X)\cong X(A)$$ naturally in $$A$$. That is, that the functors $$[\mathscr A^{\text{op}}, \textbf{Set}](H_\bullet,X)\text{ and } X(\bullet)$$ are naturally isomorphic. To see why the map $$[\mathscr A^{\text{op}}, \textbf{Set}](H_A,X)\to [\mathscr A^{\text{op}}, \textbf{Set}](H_B,X)$$ is defined in the way described by Leinster, we decompose the functor in question viz. $$[\mathscr A^{\text{op}}, \textbf{Set}](H_\bullet,X)$$ as a composition of two functors. This is implicitly done on p.95 (see the related question). The decomposition, according to p.95, is

$$\mathscr A^{op}\to[\mathscr A^{\text{op}}, \textbf{Set}]^{op}\to\mathbf {Set}\\ A\mapsto H_\bullet(A)=H_A\mapsto [\mathscr A^{op},\mathbf{Set}](H_A,X)$$

where the first functor is $$H_\bullet^{op}$$ and the second functor is $$[\mathscr A^{op},\mathbf{Set}](-,X)$$.

Here is how the composition works on arrows.

• Let $$f:B\to A$$ be an arrow in $$\mathscr A$$ (so that $$f^{op}:A\to B$$ is an arrow in $$\mathscr A^{op}$$).

• Under the functor $$H_\bullet^{op}:\mathscr A^{op}\to[\mathscr A^{\text{op}}, \textbf{Set}]^{op}$$, the arrow $$f^{op}$$ gets mapped to the arrow $$H_\bullet^{op}(f^{op})=H_\bullet(f)^{op}=H_f^{op}:H_A\to H_B$$. Here we used Definition 4.1.21 and the definition of the opposite functor.

• Then, under the functor $$[\mathscr A^{op},\mathbf{Set}](-,X): [\mathscr A^{\text{op}}, \textbf{Set}]^{op}\to\mathbf {Set}$$, the arrow $$H_f^{op}$$ gets mapped to the arrow $$-\circ H_f:[\mathscr A^{op},\mathbf{Set}](H_A,X)\to [\mathscr A^{op},\mathbf{Set}](H_B,X)$$. Here we used Definition 4.1.16.

For the second question, we are dealing with the functor $$[\mathscr A^{op},\mathbf{Set}](H_A,-)$$, which is not even a composite functor, so we can directly apply Definition 4.1.1.