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

Answer 2

$X$ is Hausdorff. Let $x_1,x_2\in X$. Since $g\circ f=1_X$, then $f$ is 1-1, and hence $y_1=f(x_1)\ne f(x_2)=y_2$. As $Y$ is a Hausdorff space, then there exist $V_1,V_2\subset $Y open and disjoint, with $y_1\in V_1$ and $y_2\in V_2$. Clearly $U_1=f^{-1}(V_1)$ and $U_2=f^{-1}(V_2)$ are open and disjoint subsets of $X$, and $x_1\in U_1$, $x_2\in U_2$.

$f(X)$ is closed. Observe that $f(X)=g^{-1}(X)$. Note that, as $g$ is continuous, then for every $K\subset X$ closed in $X$, the set $g^{-1}(K)$ is closed in $Y$.

Answer 3

The answers by Yiorgos S. Smyrlis and s.h already give very short answers for both your questions, but you might be interested in more background:

$f$ is a section (with $g$ being the corresponding retraction) and therefore $f$ is injective (while $g$ is surjective). If $Y$ is Hausdorff and we have an injective continuous maps $f\colon X\rightarrow Y$, then $X$ is Hausdorff. I gave two different proofs for this here, also using the diagonal criteria for Hausdorff spaces like you to once. You also find a proof here.

Since $g\circ f=\operatorname{id}_X$ is open and $f$ is continuous, we have that $g\vert_{f(X)}\colon f(X)\rightarrow X$ is open. I gave a proof for this here. Since it is also continuous and surjective, it is a quotient map. So you even have the more general result, that a subset $U\subseteq X$ is open iff $g\vert_{f(X)}^{-1}(U)=g^{-1}(U)\cap f(X)\subseteq Y$ is open. Considering $U=X$ and observing $g^{-1}(X)=f(X)$ would give your result as a special case. (But as Yiorgos S. Smyrlis already answered, that observation would already directly yield your result.)

If $X$ is compact, you furthermore have, that $f$ is a topological embedding (See here.) and therefore is a homeomorphism onto its image (if $f(X)\subset Y$ is equipped with the subspace topology).

Answer 4

Let us see that $X$ is Hausdorff, in effect: Let $x_1,x_2 \in X$, as $g\circ f=1_x$ then $f$ is injective and therefore. $$f(x_1)\neq f(x_2)\longrightarrow y_1\neq y_2$$ Since $Y$ is a hausdorff space then there exist $V_1,V_2 \subset Y$ open and disjoint such that $$y_1\in V_1 \ y \ y_2\in V_2$$ Since f is continuous $$f^{-1}(V_1)=U_1$$ $$f^{-1}(V_2)=U_2$$

With $U_1$, $U_2$ are open and furthermore, \begin{equation*} \begin{split} U_1\cap U_2=&f^{-1}(V_1)\cap f^{-1}(V_2)\\ =&f^{-1}(V_1\cap V_2)\\ =&f^{-1}(\emptyset)\\ =&\emptyset \end{split} \end{equation*} where $U_1, U_2 \in X$, with $x_1in U_1, x_2 \in U_2$, so $X$ is hausdorff. \vspace{5mm} \Now we prove that $f(X)$ is closed in Y, indeed, let us consider the function:

\begin{equation*} \begin{split} h: Y \longrightarrow \ X \times Y & \longrightarrow Y\times Y \\ y \longrightarrow (g(y),y) &\longrightarrow (f(g(y)),y) \end{split} \end{equation*}

$h$ is continuous because it is the composition of continuous functions, as $Y$ is hausdorff the diagonal $\Delta_y \subset Y \times Y$ is closed and therefore, $$f(X)=h^{-1}(\Delta_y)$$ is closed.