# Example of quotient map from Munkres book

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

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. – Don Thousand Oct 21 '18 at 22:53

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 $$Y$$.

• Thanks a lot for reply! Very nice and detailed explanantion! – ZFR Oct 22 '18 at 16:23
• You're welcome. Glad to help! – Matematleta Oct 22 '18 at 19:16

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

• 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 '18 at 4:28
• @HennoBrandsma, yes indeed, the notion of compactness comes after this! – ZFR Oct 22 '18 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.