Prove that a continuous inverse-transformation of $f: [0,1) \cup \{ 2 \} \to [0,1]$ exists I am having this transformation $f: [0,1) \cup \{ 2 \} \to [0,1]$ 

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

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