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

Question

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.

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

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

Answer 1

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

Answer 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?