0
$\begingroup$

I am having this transformation $f: [0,1) \cup \{ 2 \} \to [0,1]$

$$f(x) = \begin{cases} x & x \neq 2 \\1 & x = 2 \end{cases}$$

I've already proved that it is continuous.


Question: Is there a continuous inverse-transformation of this transformation?


My thoughts: Because f is continous and strictly increasing, then f has an inverse function $ f^{-1}: [0,1] \to [0,1) \cup \{ 2 \} $

$\endgroup$
2
2
$\begingroup$

Your function is a bijection, so an inverse definitely exists. You could probably write that function down by hand if you needed to.

As for continuity: One theorem about continuous functions is that the image of a connected topological space under a continuous map is also connected. What does that tell you about the continuity of the inverse function?

$\endgroup$
-1
$\begingroup$

Because f is continous and strictly increasing, then f has an inverse function $ f^{-1}: [0,1] \to [0,1[ \cup \{ 2 \} $

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.