Centers and subrings of centers are isomorphic I have been trying to get this problem right for many hours, is it correct?
Let $R$ be a ring prove that the center of $C$ is isomorphic to the center of $Z(R)$. $C$ is the set of group homomorphisms from $R$ as an abelian group to $R$.
Attempt:  For each $r \in R$, let $x_r:R \rightarrow R$ be given by $x_r(s)=sr$ for all $s \in R$. Suppose $\alpha \circ x_r=x_r \circ \alpha$ for all $r \in R$. Then for each $s \in R$, we have
$\alpha(x_r(s))=(\alpha(s))r$
Since $\Psi:R \rightarrow C$ given by $t \mapsto \alpha_t$ is injective, $R$ is isomorphic to a subring of $C$. Since $Z(C) \subset \Psi(R)$, we have that $\Psi^{-1}(C)$ is a subring of $R$. But $\Psi^{-1}(C) \subset Z(R)$. So that $\Psi^{-1}(C)$ is a subring of $Z(R)$. Thus, $Z(C)$ is isomorphic to a subring of $Z(R)$.
 A: If $f : A \to B$ is a ring homomorphism, then $\operatorname{im} f$ is a subring of $B$. In particular, if $f$ is injective, then $A$ is isomorphic to a subring of $B$.

Claim 1: If $\alpha \in Z(\operatorname{End}_\textsf{Ab}(R))$, then $\alpha(1) \in Z(R)$.

Proof: Let $r \in R$. Define $\lambda_r : R \to R$ and $\mu_r : R \to R$ by $\lambda_r(x) = rx$ and $\mu_r(x) = xr$ for all $x \in R$. Then $\lambda_r$ and $\mu_r$ they live in $\operatorname{End}_\textsf{Ab}(R)$, and because $\alpha$ is in its center, then $\alpha \circ \lambda_r = \lambda_r \circ \alpha$ and $\alpha \circ \mu_r = \mu_r \circ \alpha$. Evaluating at $1$ on both sides of the first equality we get that $\alpha(r) = r\alpha(1)$, and also $\alpha(r) = \alpha(1)r$ using the second one. Thus $r\alpha(1) = \alpha(1)r$, and the claim follows since $r$ was arbitrary. $\blacksquare$

Claim 2: Evaluation at $1$ is an injective ring homomorphism $e : Z(\operatorname{End}_\textsf{Ab}(R)) \to Z(R)$.

Proof: Let $\alpha$ and $\beta$ in $Z(\operatorname{End}_\textsf{Ab}(R))$. Then $(\alpha+\beta)(1) = \alpha(1)+\beta(1)$ tells us that $e(\alpha+\beta) = e(\alpha)+e(\beta)$, and $\alpha(r) = \alpha(1)r$ with $r = \beta(1)$ tells us that $e(\alpha \circ \beta) = e(\alpha)e(\beta)$. Also, $e(\operatorname{id}_R) = 1$, and thus $e$ is a ring homomorphism. Finally, it is injective since if $\alpha(1)=\beta(1)$, then $\alpha(r) = \alpha(1)r = \beta(1)r = \beta(r)$ for all $r \in R$, and so $\alpha = \beta$. $\blacksquare$
