Isomorphism construction and tensor product How to construct a isomorphism from $End(U \otimes V)$ to $Hom(U,U \otimes End(V))$?
It seems that I need to compose some canonical maps but I'm not sure how to actually do it.
 A: I'm not sure what setting you are working in, so I will go through this in the case of vector spaces over a field $k$, but it all generalises to (bi-)modules over appropriate rings.
The map you want can be seen as the composition of two standard isomorphisms. The first is the tensor-hom adjunction:
$$ \mathrm{Hom}(X\otimes Y,Z) \cong \mathrm{Hom}(X,\mathrm{Hom}(Y,Z)). $$
This sends a map $f\colon X\otimes Y\to Z$ to the map $\bar f\colon X\to\mathrm{Hom}(Y,Z)$ such that $\bar f(x)(y)=f(x\otimes y)$. The inverse sends a map $g\colon X\to\mathrm{Hom}(Y,Z)$ to $\hat g\colon X\otimes Y\to Z$ such that $\hat g(x\otimes y):=g(x)(y)$.
There is also a natural map
$$ Y\otimes\mathrm{Hom}(X,Z) \to \mathrm{Hom}(X,Y\otimes Z). $$
This sends $y\otimes f$ to the map $X\to Y\otimes Z$, $x\mapsto y\otimes f(x)$. This requires some finiteness condition to be an isomorphism, though. One needs that either $X$ or $Y$ is finite dimensional (or more generally either $X$ or $Y$ is finitely generated projective).
Note that this is commonly applied when $Z=k$, yielding the map
$$ Y \otimes D(X) \to \mathrm{Hom}(X,Y), \quad D(X) := \mathrm{Hom}(X,k) $$
So, putting these together, one obtains the natural map
$$ \mathrm{Hom}(U,U\otimes\mathrm{End}(V)) \to \mathrm{Hom}(U,\mathrm{Hom}(V,U\otimes V)) \to \mathrm{Hom}(U\otimes V,U\otimes V) = \mathrm{End}(U\otimes V) $$
sending $f\colon U\to U\otimes\mathrm{End}(V)$ to the endomorphism $\tilde f$ of $U\otimes V$ given by
$$ \tilde f(u\otimes v)=\sum_iu_i\otimes g_i(v), \textrm{ where }
f(u)=\sum_iu_i\otimes g_i, $$
and this will be an isomorphism whenever $U$ is finite dimensional (finitely generated projective).
