Construction of Tensor Product of Functors Suppose $D$ is cocomplete, $S:C^{op}\rightarrow Set$ and $T:C\rightarrow D$ are functors. Then, I know that the tensor product of $S$ and $T$ is defined to be the coend $\int ^{c}Sc\cdot Tc$, which exists because $C$ is cocomplete. But I have not seen an explicit construction, from scratch so I want to do it myself. 
Here is an outline of my attempt. My question is whether I am on the right track and if not, where I should be going. 
$(1)$. Define $F:C^{op}\times C\rightarrow D $ by $F(c,c')=\coprod _{i\in Sc}Tc'$. 
Define $F$ on arrows as follows:
If $f:d\rightarrow c$ and $g:c'\rightarrow d'$ then we need an arrow 
$\coprod _{i\in Sc}Tc'\rightarrow \coprod _{k\in Sd}Td'$
Consider the commuting square
\begin{matrix}Tc' &\stackrel{Tg}{\rightarrow}&Td'\\\downarrow{\beta_{Sf(k)}}&&\downarrow{\lambda _k}\\ \coprod _{\left \{ i\in Sc:Sf(k)=i \right \}}Tc'& \stackrel{\phi }{\rightarrow}&\coprod _{_{k\in Sd}}Td'\end{matrix}
where $\lambda _k$ and $\beta _{Sf(k)}$ are the usual coproduct injections and $\phi $ is the unique arrow provided by the UMP of the coproduct. 
We also have a unique $\psi :\coprod _{\left \{ i\in Sc\right \}}Tc'\rightarrow \coprod _{\left \{ i\in Sc:Sf(k)=i \right \}}Tc'$ such that $\psi \circ \beta_i=\beta_{Sf(k)}$
We set $F(f,g)=\phi \circ \psi$.
Of course, one has to show $F$ is a functor. 
$(2)$ Suppose $f:c'\rightarrow c$.  Then, with arrows $h_c:\coprod _{k\in Sc}Tc,\rightarrow \coprod _{c\in C}\left ( \coprod _{k\in Sc}Tc \right )$ the coproduct injections, we have the square
\begin{matrix}\coprod _{i\in Sc}Tc' &\stackrel{F(1,f)}{\rightarrow}&\coprod _{i\in Sc}Tc\\\downarrow{F(f,1)}&&\downarrow{h_c}\\\coprod _{k\in Sc'}Tc'& \stackrel{h_{c'}}{\rightarrow}&\coprod _{c\in C}\left ( \coprod _{k\in Sc}Tc \right )\end{matrix}.
and the following data: 
$\lambda _{k}:Tc\rightarrow \coprod _{i\in Sc}Tc$
$\beta _{k}:Tc'\rightarrow \coprod _{i\in Sc}Tc'$
$\lambda ' _{k}:Tc'\rightarrow \coprod _{k\in Sc'}Tc'$
$F(f,1)=\phi _1:\coprod _{i\in Sc}Tc'\rightarrow \coprod _{k\in Sc'}Tc$
$F(1,f)=\phi : \coprod _{i\in Sc}Tc'\rightarrow \coprod _{i\in Sc}Tc$
and the conditions:
$\phi \circ \beta _k=\lambda _k \circ Tf$ and $\phi _1 \circ \beta _k= \lambda '_k$
For the square to commute we would require 
$h_c \circ \phi =h_{c'}\circ \phi _1$ which I think means either that one must take a quotient of $C$ or find a coequalizer for a certain pair of maps. On the other hand, if $C$ is not small then $\coprod _{c\in C}\left ( \coprod _{k\in Sc}Tc \right )$ doesn't even make sense so maybe this is the wrong approach altogether. 
 A: You seem to be on the right track; I think there is no particular problem in resolving size issues embedding everything in a suitably large universe, so that all boils down to the verification of the initial cowedge condition for the family
$$
\left\{Sx \cdot Tx \xrightarrow{\omega_x} \int^c Sc \cdot Tc\right\}_{x\in \cal C}
$$
Once you are convinced of this, it is possible to take a more conceptual point of view: first of all, recall that an adjunctions of two variables is a triple $\mathfrak t=\{\otimes , \wedge, [-,=]\}$ of (bi)functors between three categories $\mathbf{S}, \mathbf{A}, \mathbf{B}$, defined via the adjunctions
$$
\hom_{\mathbf{B}}(S\otimes A, B)\cong \hom_{\mathbf{S}}(S, [A,B])\cong \hom_{\mathbf{A}}(A, S\land B).
$$ 
Such an isomorphism uniquely determines the domains ant the variance of the three functors involved, in each variable, but to be clear $\otimes \colon \mathbf{S}\times \mathbf{A}\to \mathbf{B}$, and then $\land \colon \mathbf{S}^\text{op}\times \mathbf{B}\to \mathbf{A}$, and $[-,=] \colon \mathbf{A}^\text{op}\times \mathbf{B}\to \mathbf{S}$. The prototype of this situation is the tensor, hom, cotensor functors of a suitably nice monoidal category, or the copower, hom, and power of a co/tensored enriched category.
Now, given an adjunction of three variables $\mathfrak t$, one can induce another adjunction of three variables $\mathfrak{t}' = \{\boxtimes, \curlywedge, \langle-,=\rangle \}$, on the categories $\mathbf{S}^{\mathbf{I}^\text{op}\times \mathbf{J}}, \mathbf{A}^\mathbf{I}, \mathbf{B}^\mathbf{J}$, for any $\mathbf{I}, \mathbf{J}\in\mathbf{Cat}$; to see this, start defining $F\boxtimes G\in \mathbf{B}^\mathbf{J}$ out of $F\in \mathbf{S}^{\mathbf{I}^\text{op}\times \mathbf{J}}, G\in \mathbf{A}^\mathbf{I}$, as the coend 
$$
\int^i F(i, -)\otimes Gi
$$
It is an instructive exercise in coend calculus to show that this induces suitable functors $\curlywedge$ and $\langle -,=\rangle$: for example
$$\begin{align*}
\mathbf{B}^\mathbf{J}(F\boxtimes G,H) & \cong \int_j \mathbf{B}((F\boxtimes G)j, Hj)\\
& \cong \int_j \mathbf{B}\left(\int^i F(i,j)\otimes Gi, Hj\right)\\
& \cong \int_{ij} \mathbf{B}( F(i,j)\otimes Gi, Hj)\\
& \cong \int_{ij} \mathbf{A}(Gi, F(i,j)\land Hj)\\
& \cong \int_{i} \mathbf{A}\left(Gi, \int_j F(i,j)\land Hj\right)\\
& \cong \mathbf{A}^\mathbf{I}(G, F\curlywedge H)
\end{align*}$$
if we define $F\curlywedge H\colon i\mapsto \int_j F(i,j)\land Hj$.
Do the other as an exercise!
