Let $\phi$ be an onto ring homomorphism that maps say $R$ to $S$.
We know that under $\phi$, $Z(R)$ (the center of $R$, commutative elements under multiplication in $R$) get maps to a subset of $Z(S)$. Hence $\phi\big(Z(R)\big)\subseteq Z(S)$.
I can show that this holds, but I am struggling to come up with a counterexample showing that the image of $\phi$ under $Z(R)$ does not necessary have to be the entire set $Z(S)$. Namely $\phi\big(Z(R)\big)=Z(S)$ is not always true.
Then there must be some element say $\bar{t}\in Z(R)$ (since $\phi$ is onto, $\exists t\in R$ where $\phi(t)=\bar{t}$) such that for some $r\in R$, $tr\neq rt$ but $\bar{t}s=s\bar{t}$ for all $s\in S$. In another words, there must be some that is commutative in $S$ but the pullback of that element is not commutative in the domain.
This would be simple if $\phi$ is not onto, but it is.
If there's any logical error above, or any hints/advice, I would be appreciated. Thanks.
You can take the free $\mathbb{R}$-algebra $F$ on two generators $x,y$. It is a non-commutative algebra. You can think of it as the noncommutative analogue of $\mathbb{R}[x,y]$ (see here). Consider the polynomial ring $\mathbb{R}[x,y]$ which is the free commutative $\mathbb{R}$-algebra on two generators $x,y$. Define the onto $\mathbb{R}$-algebra homomorphism $\phi:F \to \mathbb{R}[x,y]$ as the one induced by the identity on $\{ x,y\}$.
Then $Z(F)$ is given by the constant polynomials in $F$. Thus its image under $\phi$ is the subring of constant polynomials of $\mathbb{R}[x,y]$. Therefore $$\phi(Z(F))= \mathbb{R} \subsetneq \mathbb{R}[x,y] =Z(\mathbb{R}[x,y]).$$
Clearly this still works with any commutative ring in place of $\mathbb{R}$.