Let $f: X \to Y$ be a bijective continuous function. If $X$ is compact, and $Y$ is Hausdorff, then $f:X \to Y$ is a homeomorphism.
My goal is to demonstrate the necessity of both the compact and Hausdorff property of $X$ and $Y$ respectively. I want to know if my counter examples are correct:
Let $f: X \to Y$ be the identity function.
(1) Let $X=[0,1]$ with the standard topology and let $Y=[0,1]$ with the indiscrete topology. Then $f: X \to Y$ is continuous and bijective but not a homeomorphism. Since $U=(1/3,1/2)$ is open in $X$, but $(f^{-1})^{-1}(U)=f(U)$ is not open in $Y$.
(2) Let $X=Y=\mathbb{R}$. Let $Y$ have the standard topology and $X$ have the discrete topology. Let $U=\{x\}$ since open sets are singletons. $f(^{-1})^{-1}(x)=f(x)$ is closed in $Y$ since $Y$ has the standard topology. Hence $f$ is not a homeomorphism.
This is the old question
My goal is to demonstrate the necessity of both the compact and Hausdorff property of $X$ and $Y$ respectively. I want to know if my counter examples are correct:
Let $f$ be the identity function. Then $f: X \to Y$ be a bijective continuous function. If $X=[0,1]$ with the standard topology and $Y=[0,1]$ with the indiscrete topology we have that $f:X \to Y$ is not a homeomorphism.
Let $f: X \to Y$ be a bijective continuous function, $X=(0,1)$, and $Y=[0,1]$ both with the standard topology. It's not a homeomorphism because $f^{-1}([0,1])$ is not connected, but $[0,1]$ is connected.
The first shows why Hausdoff is necessary for $Y$ and the second shows compactness is necessary for $X$.
Are both of the examples correct? I have no confidence in my first example because I'm having problems showing $f^{-1}$ is not continuous. However, I am confident in my second example.
Best Answer
Second edit: Your new examples are both correct.
First edit: I wrote that the first example was incorrect. That was a typo. It is actually correct.
Your first example is correct. You need to show that $f^{-1}$ is not continuous, that is, that it is not true that $(f^{-1})^{-1}(U)=f(U)$ is an open set for all open sets $U$. Take any set $U$ which is open in $X$ but which is not open in the indiscrete topology, for example $U=(1/3,1/2)$. Then $f(U)=U$, which is not open in $Y$. Thus $f$ is not a homeomorphism.
Your second example is incorrect because there is no bijective continuous $f$ from $(0,1)$ to $[0,1]$: you must have $f(x)=0$ for some $0<x<1$, so by the intermediate value theorem, there must exist $y<x<z$ such that $f(y)=f(z)=\epsilon$ for some small $\epsilon>0$ (think of the curve of $f$ dipping down to $0$ and then going up again; it must pass through equal values somewhere on opposite sides of the zero).