Suppose $R,S$ are arbitrary rings not necessarily commutative. We define:
$Z(R) =$ {$a \in R: \forall b\in R$ we have $ab=ba$}
We must show that if $\phi: R$ -> $S$ is surjective, then $\phi(Z(R)) \subseteq Z(S)$ and if $\phi: R$ -> $S$ is injective, then it is not true.
So, this is my logic so far:
If $\phi: R$ -> $S$ is surjective, then $\forall s \in S: \exists r\in R$ where $s=\phi(r)$. This just confirms for us that there must exist at least one element in $R$ that maps to $S$.
Suppose we had $a,b \in Z(R)$, then since $\phi: R$ -> $S$ is a ring homomorphism, then multiplication must be preserved. Thus, $\phi(ab) = \phi(a)\phi(b)$ which simply means that now, both $a,b \in S$ and thus, $\phi(Z(R)) \subseteq Z(S)$.
I am not too sure of the validity of this proof and I am also not really sure how injectivity/surjectivity really affect anything.
Any help would be much appreciated.
You need surjectivity to say that whenever $\varphi(a)$ commutes with every $\varphi(b)$, then $\varphi(a)$ is in the center of $S$. Note, you should only assume that $a \in Z(R)$ and from the fact that it commutes with every $\varphi(b)$ conclude that $\varphi(a)$ is in $Z(S)$.
For a counterexample if $\varphi$ is not surjective, consider the map $\varphi: \mathbb Z \rightarrow \text{GL}(2, \mathbb Z)$ given by $$n \mapsto \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix}.$$