product of end equal to end of product In the theory of ends, is it true in general that the end of a product is equal to the product of the ends? If not, then what are sufficient conditions for it to be true?
For instance, if $F,G,H,I\in[\mathcal{C}^{\text{op}},\text{Set}]$, is it true that
$$\int_Y \text{Set}(F(Y),G(Y))\times \int_Y\text{Set}(H(Y),I(Y)) \cong
\int_Y \text{Set}(F(Y),G(Y))\times\text{Set}(H(Y),I(Y))$$
I seem to have seen steps like this done in a few places but haven't found a justification for it. Does one prove this from 'first principles' i.e. the definitions of ends and products, or is there a more immediate way to see this is true?
 A: Limits commute with limits and ends are particular kinds of limits. But you have to take into account the correct formula for this, that is, given a functor $L : {\cal I}\times{\cal J}\to \cal C$, the limits
$$\textstyle
\lim_{\cal I}\lim_{\cal J} L\quad \lim_{\cal J}\lim_{\cal I} L\quad \lim_{\cal I\times J} L
$$ are all isomorphic (even more precisely, one exists iff all three exist and they are canonically isomorphic objects).
So, in general no, it is false; the integration variable is mute, so the red $Y$ and the blue $Y$ in the product of integrals
$$
\int_{\color{red} Y} \text{Set}(F({\color{red} Y}),G({\color{red} Y}))\times \int_{\color{blue} Y}\text{Set}(H({\color{blue} Y}),I({\color{blue} Y}))
$$ have to be treated as different. The best you can say thus is that
$$
\int_{\color{red} Y} \text{Set}(F({\color{red} Y}),G({\color{red} Y}))\times \int_{\color{blue} Y}\text{Set}(H({\color{blue} Y}),I({\color{blue} Y}))\cong\int_{\color{red} Y\color{blue} Y} \text{Set}(F({\color{red} Y}),G({\color{red} Y}))\times \text{Set}(H({\color{blue} Y}),I({\color{blue} Y}))
$$ to further reduce this to be equal to
$$
\int_{\color{green} Y} \text{Set}(F({\color{green} Y}),G({\color{green} Y}))\times \text{Set}(H({\color{green} Y}),I({\color{green} Y}))
$$ you need further assumptions on $\cal C$.
