$X$ be a Hausdorff topological space , let $f:X \to X$ be a continuous map such that $f\circ f=f$ , then is $f(X)$ is closed in $X$ ?

Question

Let $X$ be a Hausdorff topological space , let $f:X \to X$ be a continuous map such that $f\circ f=f$ , then is it true that $f(X)$ is closed in $X$ ?

Answer 1

Yes. Say $(f(x_\alpha))$ is a net in $f(X)$ and $f(x_\alpha)\to y$. Since $f$ is continuous, $f(x_\alpha)=f(f(x_\alpha))\to f(y)$. So $f(y)=y$, hence $y\in f(X)$.

Or: $f(X)=\{x\in X:f(x)=x\}$.

Heh, Exercise It's clear I think where Hausdorff is used in the second argument, the one I only hinted at. Where is Hausdorff used in the first argument?

Answer 2

Yes.

Let $x\in X-f\left(X\right)$. We have $f(x)\in f\left(X\right)$ and consequently $x\neq f(x)$.

Because $X$ is Hausdorff there are neighbourhoods $U$ of $x$ and $V$ of $f(x)$ with $U\cap V=\emptyset$.

Then $U\cap f^{-1}(V)$ is a neighbourhood of $x$ and it has an empty intersection with $f\left(X\right)$

(observe: if $y\in U\cap f^{-1}(V)\cap f\left(X\right)$ then $y=f(y)\in V$ and $y\in U$, contradicting $U\cap V=\emptyset$).

Every $x\in X-f\left(X\right)$ has a neighbourhood with that property hence $f\left(X\right)$ is a closed set.