# Topological embedding on adjunction space

I'm (self) learning topology and I'm stuck on a passage of the proof of the following theorem:

let $$X \bigcup_fY$$ be the adjunction space formed by attaching Y to X along f. Let $$q: X\sqcup Y \rightarrow X \sqcup Y / \sim$$ the quotient map for the quotient topology on $$X \sqcup Y / \sim$$. Then the restriction of $$q$$ to $$X$$ is a topological embedding.

First of all I guess that in the statement X is identified with its canonical injection image in $$X \sqcup Y$$.

Secondly the book I'm studying reports $$q^{-1}(q(B)) \bigcap X = B$$ and $$q^{-1}(q(B)) \bigcap Y = f^{-1}(B)$$ for a set B closed in $$X$$. To make sure I got this right we're considering $$X$$ as a subset of $$X \sqcup Y$$ with the disjoint union topology $$\tau$$, then we're endowing $$X$$ with the subspace topology inherited by $$\tau$$ and we're selecting a closed subset B wrt this subspace topology, right? My question is how can I derive $$q^{-1}(q(B)) \bigcap X = B$$ and $$q^{-1}(q(B)) \bigcap Y = f^{-1}(B)$$.

It should boild down to the fact that $$\sim$$ is generated by $$a \sim f(a)$$ (where $$a \in A$$, a closed subset of $$Y$$, which is the domain of $$f$$) and so points in $$A$$ with the same image are identified.

I can't really go on from here, can somebody please make clear how the preceding statement is true? Thank in advance!

So lets do it formally. We have two topological spaces $$X, Y$$ and a continuous function $$f:A\to X$$ where $$A\subseteq Y$$.

When you work with the disjoint union people often do implicit steps, like $$X\subseteq X\sqcup Y$$. But what does it really mean? For that we need the definition of the disjoint union.

It goes like this:

$$X\sqcup Y:=(\{1\}\times X)\cup(\{2\}\times Y)$$

and the topology is generated by $$\{1\}\times U$$ and $$\{2\}\times V$$ for open $$U\subseteq X$$ and open $$V\subseteq Y$$. Now $$X$$ embeds to $$X\sqcup Y$$ via $$x\mapsto (1,x)$$ and $$Y$$ embeds to $$X\sqcup Y$$ via $$y\mapsto (2,y)$$. Thus we identify $$X$$ with $$X'=\{1\}\times X$$ and $$Y$$ with $$Y'=\{2\}\times Y$$.

Now the relation on $$X\sqcup Y$$ is generated by $$(2, a)\sim (1, f(a))$$ for $$a\in A$$.

Throught I will use apostrophe $$A'$$ to indicate image of $$A$$ in $$X\sqcup Y$$. Either via $$x\mapsto (1,x)$$ or $$y\mapsto (2,y)$$ map, so the apostrophe works only for subsets of either $$X$$ or $$Y$$. I hope it won't lead to confusion.

Back to the problem. Consider the quotient map $$q:X\sqcup Y\to X\sqcup Y/\sim$$. Let $$B\subseteq X$$ and consider $$B'$$ being a subset of $$X'$$ which is a subset of $$X\sqcup Y$$. You normally identify $$B'$$ with $$B$$.

What we want to show, formally, is that $$q^{-1}(q(B'))\cap X'=B'$$. So the "$$\supseteq$$" inclusion should be clear.

The other inclusion goes like this: let $$v\in q^{-1}(q(B'))\cap X'$$. This means $$v=(1,x)$$ for some $$x\in X$$ and $$v\in q^{-1}(q(B'))$$. Therefore $$v\sim w$$ for some $$w\in B'$$. By the definition of $$\sim$$ this implies that $$v=(1,b)$$ for some $$b\in B$$ or $$v=(2,a)$$ for some $$a\in A$$. But the last case is impossible since $$v=(1,x)$$. Therefore $$v=(1,b)$$ for some $$b\in B$$, i.e. $$v\in B'$$. This completes the proof. $$\Box$$

Now for the other equality. Again, we start with $$B\subseteq X$$, $$B'\subseteq X'$$ and $$Y'=\{2\}\times Y$$. Formally we want to show that

$$q^{-1}(q(B'))\cap Y'=f^{-1}(B)'=\{2\}\times f^{-1}(B)$$

Again, normally we identify $$f^{-1}(B)$$ with $$f^{-1}(B)'$$ but we want to do this formally.

So the "$$\supseteq$$" inclusion should be clear. If $$b\in f^{-1}(B)$$ then obviously $$(2,b)\in Y'$$. On the other hand since $$f(b)\in B$$ then $$(2,b)\sim v$$ for some $$v\in B'$$, precisely for $$v=(1, f(b))$$. Meaning $$(2,b)\in q^{-1}(q(B'))$$. This completes this inclusion. $$\Box$$

For the "$$\subseteq$$" inclusion consider element $$w\in q^{-1}(q(B'))\cap Y'$$. Since $$w\in Y'$$ then $$w=(2,y)$$ for some $$y\in Y$$. On the other hand $$w\in q^{-1}(q(B'))$$ meaning $$w\sim v$$ for some $$v\in B'$$. But $$v=(1,b)$$ for some $$b\in B$$. So we have this situation

$$(2,y)\sim (1,b)$$

By the definition of $$\sim$$ it follows that $$b=f(y)$$ and so $$y\in f^{-1}(b)$$. In particular $$w\in f^{-1}(B)'$$ which completes the proof. $$\Box$$

Final note: So I get it that this requires lots of other symbols, this weird $$(1,x), (2,y)$$ products and proofs/constructions become long. But that way we solve all possible problems and we avoid issues like: what is $$X\sqcup X$$? How can we distinguish one embedding $$X\subseteq X\sqcup X$$ from another? Once we grasp formalities we can be sure that our intuition leads us in a correct direction.