I would like to check if what I'm doing is correct or if I'm missing something:
Given the two sets $$X = [0,1] \cup (2,3], \quad Y = [0,2],$$ both equipped with the standard topology, consider the function $f:X \to Y$ defined by $$f(x) = \begin{cases} x & \text{if $x \in [0,1]$} \\ x-1 & \text{if $x \in (2,3]$} \end{cases}.$$
Show that $f$ is bijective from $X$ to $Y$ and continuous, but that $f^{-1}$ is not continuous.
- To show that $f$ is continuous, I take the open interval $V = (0,2)$ and find that $\operatorname{Preim}_f(V) = (0,1) \cup (2,3)$, which is open since it's the union of two open sets.
- To see that $f$ is bijective, I find the map $$g(x) = \begin{cases} x & \text{if $x \in [0,1]$} \\ x+1 & \text{if $x \in (1,2]$} \end{cases},$$ which is the inverse of $f(x)$. Since I found an inverse, then $f$ is bijective.
- Now, what open subset can I consider in order to show that $g = f^{-1}$ is not continuous?
Are the first 2 parts complete, or am I missing anything?
Thanks in advance!
Best Answer
Your first part is incomplete. All you have shown is that there exists one open subset of the codomain such that its preimage is open.
However, you have to show that that is the case for all open subsets of $[0, 2]$.
(The pasting/glueing lemma may be helpful here.)
To show that the inverse is not continuous, consider the open set $\left(\dfrac12, 1\right] \subset X$.
(Why is this open?)