I am having trouble verifying the following (this is self-study):
There is an isomorphism between the center of a ring $A$ and the ring of endomorphisms of the identity functor of the category of (right) $A$-modules.
The map $\Psi\ \colon Z(A) \to \mathrm{End}(1_{Mod\ A})$ sends $a \in Z(A)$ to multiplication by $a$ (let's write $\Psi(a) = \theta^{a}$). This poses no trouble.
It seems reasonable to define the inverse map $\Xi\ \colon \mathrm{End}(1_{Mod\ A}) \to Z(A)$ as sending $ \eta \in \mathrm{End}(1_{Mod\ A})$ to $\eta_A(1)$, since from this we readily get $\Xi \circ \Psi (a) = \Xi(\theta^a)=\theta^{a}_{A}(1)=1a=a$.
However I can't see why $\Psi \circ \Xi = 1_{\mathrm{End}(1_{Mod\ A})}$. Unfolding, we have to prove that $\Psi \circ \Xi (\eta) = \theta^{\eta_A(1)} = \eta$, that is, for any right $A$-module $L$, we need to have $\theta^{\eta_A(1)}_L = \eta_L$. I'm at a loss on this. I tried to find the right components on which to use naturality (that's how I succeeded to prove that $\eta_A(1)$ does lie in $Z(A)$, which wasn't obvious to me from the outset), but to no avail.
Some googling led me to this blog post by Q. Yuan (see the "sub-example") where the direction of the map $\Xi$ is deduced from the fact that $A$ is a generator of $\mathrm{Mod}\ A$. However I can't quite see explicitly why (in the perspective adopted there, I believe this is just rephrasing the issue I'm having) the lifting of a central element to an endomorphism of $1_{\mathrm{Mod}\ A}$ is inverse to the composition of the two isomorphisms with the natural injection $\mathrm{End}(1_{\mathrm{Mod}\ A}) \to Z(\mathrm{End}_A(A))$.