Can you pull a constant out of a coend? Let $F$ be a functor $\mathscr{C}^\text{op}\times\mathscr{C}\to\mathbf{Set}$, and let $S$ be an arbitrary set. Can we write the following?
$$
\int^{C:\mathscr{C}} S\times F(C,C) \cong S\times\int^{C:\mathscr{C}} F(C,C)
$$
This seems intuitively reasonable by analogy to integrals, and it would be generally useful in proofs. Is it true and if so how can I prove it?
(edited: of course it should be $\cong$, not $=$.)
 A: The coend $\int^c F(c,c)$ is a colimit in a category $D$, so every left adjoint functor $L : D\to E$ (a particular example of which is $S\times-$ in a cartesian closed category) will preserve such colimit, which means that provided the colimit exists in the codomain of $L$, you have $L\left(\int^c F(c,c)\right)\cong \int^c LF(c,c)$. So, for example,
$$
S\times \int^c F(c,c)\cong \int^c S\times F(c,c)
$$ for every functor $F : C^o\times C\to Set$ and set $S$; but also
$$
S\otimes \int^c F(c,c)\cong \int^c S\otimes F(c,c)
$$ for every functor $F : C^o\times C\to D$ and object $S\in D$ where $D$ is a monoidal closed category (pointed spaces, modules over a ring,..); but also (since the functor $List$ that sends a set to its "Kleene star", being the free monoid functor, is a left adjoint)
$$
List\left(\int^c F(c,c)\right)\cong \int^c List(F(c,c))
$$ whenever (for example) $F : FinSet^o\times FinSet \to Set$ is a functor (I take finite sets to have a small category of indices for $F$); but also,
$$
\pi_0\left(\int^c F(c,c)\right)\cong \int^c \pi_0(F(c,c))
$$ for every functor $F : C^o\times C \to Spaces$, where $Spaces$ is a decent category of topological spaces and $\pi_0$ takes connected components.
A: Yes, this is true. It can be proved in several way. Let's see that in a concrete way. A point of $∫^c S×F(c,c)$ is given by some triple $(c,s,x)$ with $\newcommand{\C}{\mathscr{C}}c∈\C$, $s∈S$ and $x ∈ F(c,c)$. For each $f : c→c'$ in $\C$, $s ∈ S$ and $x ∈ F(c',c)$, we impose the relation $(c,s,F(f,1)(x)) = (c',s,F(1,f)(x))$.
$\require{AMScd}$
\begin{CD}
@>>> c\\
{} @VV{f}V\\
@. c' @>>>
\end{CD}
From this description, we see that it results in $S$ copies of $∫^c F(c,c)$.
Another possibility is to use the universal property of the coend:
$$\begin{align*}\newcommand{\Hom}{\operatorname{Hom}}\Hom\left(∫^c S×F(c,c), X\right) &≅ ∫_c \Hom(S×F(c,c),X)\\
&≅ ∫_c \Hom(F(c,c),X^S)\\
&≅ \Hom\left(∫^c F(c,c),X^S\right)\\
&≅ \Hom\left(S×∫^c F(c,c),X\right)\end{align*}$$
Or we can write the coend as a colimit and use the fact that products distribute over colimits in sets.
