Is there a counter example?

Question

Is this true? If $f$ and $g $ are continuous functions , and if $f\circ g$ is closed(open) , neither $g$ nor $f$ is necessarily closed(open)?

Answer 1

Let $f:I\hookrightarrow \Bbb R$ be the inclusion of the unit interval into the real line, and let $g:\Bbb R \twoheadrightarrow I$ be the retraction sending $(-\infty,0]$ to $\{0\}$ and $[1,\infty)$ to $\{1\}$. Then $gf$ is a homeomorphism $I\approx I$, thus open, but neither $f$ nor $g$ are open.

• $f$ maps the open $I$ to the non-open $I.$
• $g$ maps the open $(-2,-1)$ to the non-open $\{0\}.$

However, if $gf$ is open, $g$ and $f$ are continuous, and ...

• ... if $g$ is injective, then $f$ must be open.
• ... if $f$ is surjective, then $g$ must be open.

For an explanation see Under what conditions $f$ and $g$ are open maps?