Continuous function and Hausdorff space

The question is:

$f: X \to Y$ is continuous bijection, which of the following is correct:

I. if $X$ is Hausdorff space then $Y$ is Hausdorff space.

II. if $X$ is compact and $Y$ is Hausdorff space, then $f^{-1}$ exist.

I think I is correct since $X$, the Hausdorff is separable, then image should be separable. I'm not sure about II.

Besides, can you show me some example and counterexample of Hausdorff space? Thank you.

Solution 1:

Here is what is probably the simplest possible counterexample to (I).

Let $X=\{0,1\}$ with the discrete topology $\tau_d$: the open sets are $\varnothing,\{0\},\{1\}$, and $X$ itself. Let $Y=\{0,1\}$ with the indiscrete topology $\tau_i$: the only open sets are $\varnothing$ and $Y$. Define $f:X\to Y$ by $f(0)=0$ and $f(1)=1$. Then $f$ is a continuous bijection from the Hausdorff space $\langle X,\tau_d\rangle$ to the non-Hausdorff space $\langle Y,\tau_i\rangle$.

If you want something a bit bigger, let $\tau_c$ be the cofinite topology on $\Bbb R$: a set $U\subseteq\Bbb R$ is open if and only if $U=\varnothing$ or $\Bbb R\setminus U$ is finite. This is not a Hausdorff topology: if $U$ and $V$ are non-empty open sets, $U\cap V\ne\varnothing$. Now let $f$ be the identity map on $\Bbb R$, considered as a function from $\langle\Bbb R,\tau_e\rangle$ to $\langle\Bbb R,\tau_c\rangle$, where $\tau_e$ is the usual (Euclidean) topology on $\Bbb R$; then $f$ is a continuous bijection from the Hausdorff space $\langle\Bbb R,\tau_e\rangle$ to the non-Hausdorff space $\langle\Bbb R,\tau_c\rangle$.

(II), on the other hand, is true. (Since $f$ is a bijection, it’s automatic that $f^{-1}$ exists as a function; I’m assuming that you mean that $f^{-1}$ is continous.) To show this, let $U$ be any open set in $X$; we must show that $\left(f^{-1}\right)^{-1}[U]$ is open in $Y$, i.e., that $f[U]$ is open in $Y$. Let $K=X\setminus U$; since $f$ is a bijection, $f[K]=Y\setminus f[U]$. Thus, it suffices to show that $f[K]$ is closed, for then $f[U]=Y\setminus f[K]$ must be open.

$K$ is closed, since it’s the complement of an open set in $X$, and $X$ is compact, so $K$ is compact. The continuous image of a compact set is compact, so $f[K]$ is a compact set in $Y$. Finally, every compact set in a Hausdorff space is closed, so $f[K]$ is closed in $Y$, $f[U]$ is open, and $f^{-1}$ is continuous.