I'm reading the Munkres' "Topology", Section 22, Example 1.
Example 1. Let $X$ be the subspace $[0,1]\cup[2,3]$ of $\mathbb R$, and let $Y$ be the subspace $[0,2]$ of $\mathbb R$. The map $p:X\to Y$ defined by $$p(x)=\begin{cases}x&\text{for }x\in[0,1]\\x-1&\text{for }x\in[2,3]\end{cases}$$ is readily seen to be surjective, continuous, and closed.
the graph of p(Why is it presented dim and black? I don't know)
Surjectivity is just trivial.
Continuity goes like this. Consider an open interval $I$ in $Y=[0,2]$. Suppose $I$ doesn't contain $1$, w.l.o.g, suppose $I\subset[0,1]$. It follows from the monotonicity of p in [0,1] that $p^{-1}(I)$ is open. Suppose not, i.e. $I$ contains $1$. Then $p^{-1}(I)$ is an intersection of an open interval in $\mathbb R$ and $X$, thus $p^{-1}(I)$ is closed in $X$.
But I'm having difficulty in the problem of closedness of $p$.
Summary :
- Does the continuity proof right?
- How can I prove that $p$ is closed?
If $C$ is closed subset in $[0,1]\cup[2,3]$ then $C$ is the union of compact sets $C_1=C\cap[0,1]$ and $C_2=C\cap[2,3]$ and $$ p(C)=p(C_1)\cup p(C_2)=f(C_1)\cup h(C_2)$$ where $f:[0,1]\rightarrow[0,1]$ is $f(x)=x$ and $h:[2,3]\rightarrow[1,2]$ is $h(x)=x-1$. Both functions are continuous and so $f(C_1)$ is compact subset (and hence closed) of $[0,1]$ and $h(C_2)$ is a compact subset (and hence closed) of $[1,2]$. This means that $p(C)$ is closed.