Confusion about the functoriality of $H_A:\mathscr A^{op}\to \mathbf{Set}$

Question

For a locally small category $\mathscr A$, the functor $H_A:\mathscr A^{op}\to \mathbf{Set}$ is defined as follows:

• for objects $A\in\mathscr A$: $H_A(B)=\mathscr A(B,A)$
• for arrows $g:B'\to B$ in $\mathscr A$: $H_A(g):\mathscr A(B,A)\to\mathscr A(B',A),p\mapsto p\circ g$

I'm still confused with the $^{op}$ thing. To say that $H_A$ is a functor is to say that $H_A(g\circ f)=H_A(g)\circ H_A(f)$ where $g:B'\to B, f:B''\to B'$. But the composition $H_A(g)\circ H_A(f)$ doesn't make sense. What's wrong with my application of the definition of a functor?

Answer 1

I think my confusion is caused by the fact that Leinster (from whose book I borrowed the definition) uses the same letter for an arrow and the arrow it corresponds to in the opposite category. First, he considers an arrow $g:B'\to B$ in $\mathscr A$. Then, on the left-hand side of $$H_A(g):\mathscr A(B,A)\to\mathscr A (B',A)\\ p\mapsto p\circ g$$

(i.e., in $H_A(g)$), $g$ no longer refers to the arrow $B'\to B$ in $\mathscr A$ but instead it refers to the corresponding arrow $B\to B'$ in $\mathscr A^{op}$ (otherwise the notation $H_A(g)$ wouldn't make sense). And then when he writes $p\mapsto p\circ g$, his $g$ again refers to the arrow $B'\to B$.

To make things more precise, one could re-define $H_A$ on arrows in this way. If $g:B'\to B$ is an arrow in $\mathscr A$, then $$H_A(g^{op}):\mathscr A(B,A)\to\mathscr A (B',A)\\ p\mapsto p\circ g$$

So suppose we have arrows $f:B''\to B'$ and $g:B'\to B$ in $\mathscr A$. Functoriality means that $$H_A(f^{op}\circ_{op}g^{op})=H_A(f^{op})\circ H_A(g^{op}).$$ By definition, we have:

• $H_A(g^{op}): \mathscr A(B,A)\to\mathscr (B',A), p\mapsto p\circ g$
• $H_A(f^{op}): \mathscr A(B',A)\to\mathscr (B'',A), q\mapsto q\circ f$
• $H_A(f^{op}\circ_{op}g^{op}): \mathscr A(B,A)\to\mathscr (B'',A), p\mapsto p\circ (f^{op}\circ_{op}g^{op})^{op}=p\circ(g\circ f)$
• $H_A(f^{op})\circ H_A(g^{op}): \mathscr A(B,A)\to\mathscr A(B',A)\to\mathscr (B'',A), p\mapsto p\circ g\mapsto (p\circ g)\circ f$

Thus $H_A(f^{op}\circ_{op}g^{op})=H_A(f^{op})\circ H_A(g^{op})$.

Answer 2

The confusion is that you defined $H_A$ on $\mathscr A$, whereas it is defined on the opposite category.

So if $g: B\to B'$ is an arrow in $\mathscr A^{op}$ (it corresponds to $g:B'\to B$ in $\mathscr A$), you get a map $H_A(B)\to H_A(B')$

Similarly for composition, if $g:B\to B'$, $f:B'\to B''$ in $\mathscr A^{op}$, (they correspond to your $g:B'\to B, f: B''\to B'$ in $\mathscr A$), everything makes sense

It could help to look up the definition of contravariant functor (which isn't necessary as it is subsumed by the definition of functor, by just taking opposite categories, but if you're not at ease with them yet, it can be interesting)