Let $f:X\rightarrow Y$ and $g:Y\rightarrow X$ be a $\bigvee$-morphism and a $\bigwedge$-morphism of complete lattices, respectively. If $f\circ g\circ f=f$ and $g\circ f\circ g=g$, then it is true that $g\circ f\geq\mathrm{Id}_X$ and $f\circ g\leq\mathrm{Id}_Y$?
I know that the converse is true and "$g\circ f\geq\mathrm{Id}_X$ and $f\circ g\leq\mathrm{Id}_Y$" is equivalent to "$f(x)\leq y$ if and only if $x\leq g(y)$".
I tried to consider $x\in X$ and to prove that $x\leq g(f(x))$ by considering the infimum $a$ of all $t\in\mathrm{Im}(g)$ such that $x\leq t$ and obtained that $a\in\mathrm{Im}(g)$ and $x\leq a$, also I tried to use the fact that $g(f(a))=a$.
No. For instance, take a set $A$ and a map $\varphi:A\to A$ such that $\varphi\circ\varphi=\varphi$ (for instance, a constant map). Then taking $X=Y=\mathcal{P}(A)$ and $f=g=\varphi^{-1}:\mathcal{P}(A)\to \mathcal{P}(A)$, we have $fgf=f$ and $gfg=g$. However, we only have $gf\geq Id$ and $fg\leq Id$ if $\varphi$ happens to be the identity map on $A$.