Prove that $q$ is an identification function.

Question

I have problems with this exercise

Let $q:X\longrightarrow{Y} $ a continuous function. Suppose there exists a continuous function $f: Y \longrightarrow{X} $ such that $q \circ{} f$ is the identity function in $Y$. Prove that $q$ is an identification function.

I need a hint

Thanks

Feels wrong. Let $X=Y=\mathbb{R}^{+*}$ and $q=f=\frac{1}{x}$. — Jujustum, Jun 02 '21 at 20:14
An identification is an application $f: (X, \tau_X) \longrightarrow{} (Y, \tau_Y)$ surjective, such that $V \in \tau_Y$if and only if $f^{−1} (V) \in \tau_X$ , i.e, $f$ is surjective and $\tau_Y = \tau_{fin {f }} $. — Manuel Alejandro Neira Altamir, Jun 02 '21 at 21:18

score 2 · Accepted Answer · answered Jun 02 '21 at 22:02

The proof writes itself: suppose $V \in \tau_Y$. Then $q^{-1}[V] \in \tau_X$ by continuity of $q$.

The existence of $f$ implies that $q$ is surjective: $f(y)$ is a $q$ pre-image for $y \in Y$, as $q(f(y))=y$.

If $q^{-1}[V] \in \tau_X$, then $f^{-1}[q^{-1}[V]] \in \tau_Y$ by continuity of $f$, and $$f^{-1}[q^{-1}[V]] = (q \circ f)^{-1}[V] = 1_Y^{-1}[V]=V$$ and so $V \in \tau_Y$, as required.

Prove that $q$ is an identification function.

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