# Show that the restriction map $f: \ X \to Y$ is not a quotient map.

Let $$X=[0,1] \cup (2,3]$$ and $$Y=[0,2]$$.

Define a map $$f:X \to Y$$, where the toplogy on $$X$$ and $$Y$$ are the subspace topology from Euclidean topology on real line which is defined by \begin{align*}f(x)&=x, \text{ if } x \in [0,1] \\ &= x-1, \text{ if } x \in (2,3]. \end{align*}

Show that the restricted map $$f|_{[0,1] \cup (2,3)}: \ X \to Y$$ is not a quotient map.

We have $$f([0,1])=[0,1] \subseteq [0,2]$$ .
Clearly $$[0,1]$$ is open in $$X$$ with respect to subspace topology but its image $$[0,1]$$ is not open in $$Y$$.
So $$f$$ is not an open map and hence not quotient map.
An open map is a necessarily a quotient map, but the converse is not true, so your argument won't work. Use the definition: $$f$$ is a quotient map if it satisfies the condition: $$f^{-1}(A)$$ is open in $$X$$ if and only if $$A$$ is open in $$Y.$$ Now, note that $$[0,1]$$ is not open in $$Y$$ but $$f^{-1}([0,1])$$ is open in $$X$$, so $$f$$ is not a quotient map.