Let $f : (S, d_S) → (T, d_T )$ be a bijective continuous function of metric spaces, and $f^{−1}: (T, d_T ) →(S, d_S)$ be the inverse function.
Which of the following statement(s) is/are true?
$a. f^{−1}$ is always a continuous function.
$b. f^{−1}$is continuous if $(S, d_S)$ is compact.
$c. f^{−1}$ is continuous if $(T, d_T )$ is compact.
$d. f^{−1}$ is continuous if $(S, d_S)$ is connected.
My works : my answer will be option b),c) and d)
because continuous image of compact/connected is compact/connected and option a) is false take $f(x) = 1/x$
Is my answer correct ?
thanks u
$f=1/x$ from where to where and which metric to what?
For the first one, consider $f:\Bbb{R}_{dis} \rightarrow \Bbb{R}_u$ by $f(x)=x$, where $\Bbb{R}_{dis}$ denotes $\Bbb{R}$ with discrete metric and $\Bbb{R}_{u}$ denote $\Bbb{R}$ with the usual metric .Then $f$ is a continuous bijection, but $f^{-1}$ is not continuous
For the fourth and third one, consider $f:[0, 2 \pi) \rightarrow S^1$ by $f(x)=(\cos x,\sin x)$. Then $f$ is a continuous bijection but $f^{-1}$ is not continuous (at $(1,0))$ even though domain in connected and $S^1$ is compact!
The right answer is option b.
Proof: It is enough to show $f$ is open with the assumption $S$ is compact.
So let $V$ be open in $S$ and so $V^c$ is closed in $S$ and hence compact. Thus, $f(V^c)$ is a compact subset of $T$ and so closed in $T$. Since $f$ is a bijection, $$f(V)=[f(V^c)]^c$$ Hence $f(V)$ is open in $T$!