Let $f \colon X \to Y$ and $g \colon Y \to X$ with $g\circ f = 1_X$. If $Y$ is Hausdorff, then so is $X$, and $f(X)$ is closed in $Y$

Question

Let $f \colon X \to Y$ and $g \colon Y \to X$ be continuous maps with $g\circ f = 1_X$. Prove that if $Y$ is Hausdorff, then so is $X$, and $f(X)$ is closed in $Y$.

Consider the map
\begin{align}
h \colon Y &\to X \times Y \to Y \times Y \\
y &\mapsto (g(y),y) \mapsto (f(g(y)),y)
\end{align}
which is continuous because is the composition of continuous maps.
Since $Y$ is Hausdorff, the diagonal $\Delta_Y \subseteq Y \times Y$ is closed.
Therefore $f(X)=h^{-1}(\Delta_Y)$ gives the conclusion.

Is it correct?
And, how do I prove that $X$ is Hausdorff?

Answer 1

Since $f$ is injective continuous, $X$ is Hausdorff. See Pre-image of Hausdorff space is Hausdorff