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

Question

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.

Answer:

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.

Am I right?

Answer 1

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.