algebra of endomorphisms of a functor Let $C$ be a finite $k$-linear abelian category. Let $F:C \to \mathrm{Vec}$ be exact faithful functor to the category of (finite dimensional (need?)) vector spaces.  Let $\mathrm{End}(F)=\mathrm{Nat}(F,F)$ be the algebra of functorial endomorphism of $F$.
For two exact faithful functors $F_1, F_2: C \to \mathrm{Vec}$, we define $F_1\otimes F_2 : C \times C \to \mathrm{Vec}$ by $(F_1 \otimes F_2)(X, Y)= F_1(X) \otimes F_2(Y)$.
Define a morphism $\alpha_{F_1, F_2}: \mathrm{End}(F_1)\otimes \mathrm{End}(F_2) \to \mathrm{End}(F_1\otimes F_2)$ by
$$\alpha_{F_1, F_2}( \eta_1 \otimes \eta_2) |_{F_1(X)\otimes F_2(Y)}=\eta_1|_{F_1(X)} \otimes \eta_2|_{F_2(Y)}.$$
I would like to show that this is an isomorphism between $\mathrm{End}(F_1)\otimes \mathrm{End}(F_2)$ and $\mathrm{End}(F_1\otimes F_2)$.
I tried to construct an inverse morphism as follows.
Take $\eta \in \mathrm{End}(F_1\otimes F_2)$. Define $\eta_1|_{F(X)}=\eta|_{F(X)\otimes F(1)}$ and $\eta_2|_{F(Y)}=\eta|_{F(1)\otimes F(Y)}$ for objects $X, Y$ in $C$.
Using the isomorphism $F(1)\cong \mathbb{C}$, we can regard $\eta_1$ and $\eta_2$ as elements of $\mathrm{End}(F_1)$ and $\mathrm{End}(F_2)$, respectively.
Then, I want to show that $\eta=\eta_1\otimes \eta_2$ but I don't know how.
I appreciate any help.
This is Proposition 1.18.3 of the paper Tensor Categories by Etingof, Gelaki, Nikshych, and Ostrik and proving this is Exercise 1.18.4.
 A: I think your attempt to construct an inverse does not work. Already in your definition I encounter some problems. You write: 

Take $\eta \in \mathrm{End}(F_1\otimes F_2)$. Define $\eta_1|_{F(X)}=\eta|_{F(X)\otimes F(1)}$ and $\eta_2|_{F(Y)}=\eta|_{F(1)\otimes F(Y)}$ for objects $X, Y$ in $C$.

In your question you do not assume that $C$ is a monoidal category. It seems that what you denote by 1 here should be a unit objects in a monoidal category (afterwards you make use of an isomorphism $F(1) \cong \mathbb{C}$). It is not clear to me what 1 should be here if $C$ is not monoidal.
Next, the X-component of $\eta_1$ should be an endomorphism of $F_1(X)$, not of $F(X)$. Similarly, components of $\eta$ are endomorphisms of $F_1(X) \otimes F_2(X)$, not of $F(X) \otimes F(Y)$ (in your definition you write $F(X) \otimes F(1)$, maybe you are implicitly assuming $F_1=F_2=F$?).
Instead, I propose the following:
For all objects $X$ and $Y$ of $C$ there is an isomorphism of vector spaces
$$\mathrm{End}(F_1(X) \otimes F_2(Y)) \cong \mathrm{End}(F_1(X)) \otimes \mathrm{End}(F_2(Y)).$$
(I assume that $F_i(X)$ and $F_i(Y)$ are finite dimensional.)
If $\eta$ is a natural endomorphism of $F_1 \otimes F_2$, its components $\eta_{X,Y}$ are elements of $\mathrm{End}(F_1(X) \otimes F_2(Y))$.
Via the above isomorphism we obtain an element $\sum_i \eta^i_{1,X} \otimes \eta^i_{2,Y} \in \mathrm{End}(F_1(X)) \otimes \mathrm{End}(F_2(Y))$.
Now one has to check that for every $i$ the $\eta^i_{1,X}$ are the components of a natural endomorphism $\eta^i_1$ of $F_1$ and the $\eta^i_{2,Y}$ are the components of a natural homomorphism $\eta^i_2$ of $F_2$.
This way we obtain an element in $\mathrm{End}(F_1) \otimes \mathrm{End}(F_2)$ and thus a morphism
$$ \mathrm{End}(F_1 \otimes F_2) \to \mathrm{End}(F_1) \otimes \mathrm{End}(F_2), $$
which should be the inverse of $\alpha_{F_1,F_2}$.
Like Najib Idrissi I also think that you need that your functors have their images in the category of finite dimensional vector spaces, since the isomorphism I fundamentally use in the above argument does not hold in general.
