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

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

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

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 $f^{-1}: [0,1] \to [0,1) \cup \{ 2 \}$

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