I was reading the notion of quotient map, topology and space but ran into the following example.enter image description here

In this example I have understood almost everything except one moment: How to prove rigorously that $p(x)$ is closed map.

Would be thankful if anyone will show the rigorous proof.

  • It's obvious that both subfunctions of $p(x)$ are closed maps. Proving that the whole function is closed is simply a generalization of that. – Rushabh Mehta Oct 21 at 22:53
up vote 3 down vote accepted

If you want to do it from scratch, note that $f:x\mapsto x$ and $g:x\mapsto x-1$ are closed maps on $\mathbb R$ (this is easy to prove).

Let $C$ be closed in $X$, so there is a closed set $C'\subseteq \mathbb R$ such that

$X\cap C'=[0,1]\cap C'\sqcup [2,3]\cap C'=C.$ Then,

$p(C)=f([0,1]\cap C'))\sqcup g([2,3]\cap C'))=$

$[0,1]\cap C'\sqcup [1,2]\cap g(C')=[0,2]\cap (C'\cup g(C'))$.

Since $C'$ is closed in $\mathbb R$ by assumption and $g(C')$ is closed in $\mathbb R$ also, by the first remark, we conclude that $p(C)=[0,2]\cap (C'\cup g(C'))$ is closed in $Y$.

  • 1
    Thanks a lot for reply! Very nice and detailed explanantion! – RFZ Oct 22 at 16:23
  • You're welcome. Glad to help! – Matematleta Oct 22 at 19:16

Any closed subset of $[0,1] \cup [2,3]$ is compact. Since $p$ is continuous its image is compact, hence closed.

  • 1
    IIRC compactness hasn't been covered yet in the book at this point. So it's not argument Munkres had in mind probably. – Henno Brandsma Oct 22 at 4:28
  • @HennoBrandsma, yes indeed, the notion of compactness comes after this! – RFZ Oct 22 at 15:37

The map $p$ restricted to each interval is just a homeomorphism so closed. From this we conclude that this combined map is also closed.

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