Let $\phi:S\rightarrow T$ be a (homo)morphism from a regular semigroup $S$ into a semigroup $T$. Then $\textrm{im}(\phi)$ is regular. If $f$ is an idempotent in $\textrm{im}(\phi)$ then there exists an idempotent $e\in S$ such that $e\phi=f$.
This is stated without proof on page 57 of J.M. Howie's Fundamentals of Semigroup Theory.
Attempt at a proof: Since $S$ is regular, we know that $\forall x\in S, \ \exists y\in S$ such that $x=xyx$. Thus, using the fact that $\phi$ is a homomorphism: $$x\phi=(xyx)\phi=(x\phi)(y\phi)(x\phi).$$ So, $\textrm{im}(\phi)$ is regular since $\forall\ x\phi\ \exists\ y\phi\in\textrm{im}(\phi)$ so that $x\phi=x\phi y\phi x\phi$.
$\textrm{im}(\phi)=\lbrace x\phi\ \mid\ x\in S\rbrace.$ If $f$ is an idempotent element of $\textrm{im}(\phi)$ then $f=x\phi$ such that $f^2=(x\phi)^2=(x\phi)(x\phi)=x^2\phi=x\phi=f$. Therefore, let $x=e\in E_{S}$.
I'm not sure if this is correct. I feel like I've bodged it a little bit. Any help would be appreciated.
First, every $x$ has an inverse $y$, satisfying both $xyx=x$ and $yxy=y$.
(It follows from $xtx=x\,$ by taking $\,y:=txt$.)
Now, suppose $x^\phi=f=f^2$, and consider an inverse $y\ $ for $\underline{x^2}$, and set also $g:=y^\phi$. Then $$e:=xyx$$ is an idempotent: $e^2=xyx^2yx=xyx=e$ with image $f$: $$e^\phi=(xyx)^\phi=fgf=f^2gf^2=(x^2yx^2)^\phi=(x^2)^\phi=f\,.$$