# Adjunction space is a pushout

I would like to show that the diagram

$$\begin{array}{} A & \stackrel{f}{\longrightarrow} & Y \\ i \downarrow & & \downarrow {\phi_2} \\ X & \stackrel{\phi_1}{\longrightarrow} & X \coprod_f Y \end{array}$$

where $i:A \to X$ is an inclusion is a pushout.

Here $A$ is a closed subset of $X$, all maps given are continuous and $X \coprod_f Y$ is the disjoint union $X \coprod Y$ quotient by the equivalence generated by $\{(a,f(a)) \in (X \coprod Y) \times (X \coprod Y): a \in A\}$ (call the equivalence relation $\sim$)

So I start with $\nu: X \coprod Y \to X \coprod_f Y$ (the natural map) and define $\phi_1 = \nu | X$, and $\phi_2 = \nu | Y$.

Then for $a \in A$, $\phi_1(i(a)) = \nu(i(a)) = \nu(f(a)) = \phi_2(f(a))$ and the diagram is commutative.

The other part is to show that this is unique. So let $Q$ be another space such that there exists $\alpha_1:X \to Q$, $\alpha_2:Y \to Q$. We seek a $u: X \coprod_f Y \to Q$

Define the function $\Theta:X\coprod_f Y \to Q$ with $\Theta | X = \alpha_1$ and $\Theta | Y = \alpha_2$.

Then for $a \in A$, $\Theta(i(a)) = \alpha_1(i(a)) = \alpha_2(f(a)) = \Theta(f(a))$.

So this means that $\Theta$ maps elements of the equivalence class $\sim$ from $X \coprod_f Y \to Q$ (maybe I am not saying the last bit clearly, but I think it is clear what I mean!)

Is this all reasonable? I only ask because this whole commutative diagram thing is very new to me...

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No kind of hypotesis on $X$ and $Y$, right? I saw something similar studying Riemann surfaces but... I can't remember if it holds true in general top spaces. – Fosco Loregian Mar 19 '11 at 12:51
@tetrapharmakon: No (this is all in the category of topological spaces). This is building up to CW complexes – Juan S Mar 19 '11 at 12:55
This is all right (though when mapping out of the push-out you should note that $\alpha_1, \alpha_2$ are required to be equal when pulled back to $A$); the point is that the two maps give you a map of spaces $X \sqcup Y \to Q$, which factors as a map of sets $X \sqcup_f Y \to Q$. But the universal property of the quotient space states that the map $X \sqcup_f Y \to Q$ is continuous with the quotient topology. – Akhil Mathew Mar 19 '11 at 13:41
@Akhil - thank you. If you want to add as an answer, I will accept – Juan S Mar 19 '11 at 13:48

Yes, that's right. On the generating set of $\sim$ we have that $(a,0) \sim (f(a), 1)$ (where the disjoint union I denote with $0$ as the second coordinate for elements of $X$ and $1$ for elements of $Y$) and these yield the same result: either we map under $u$ to $\alpha_1(a) = \alpha_1(i(a))$ under the requirement that $u$ must commute with $\alpha_1$ or to $\alpha_2(f(a))$ under the requirement that $u$ must commute with $\alpha_2$, if we take the second guise. But by commutativity of the diagram with $i$,$f$,$\alpha_1$ and $\alpha_2$, these 2 give the same value in $Q$.
So on the generating set we have no conflict in the definition of $u$, so now you must show that $u$ is well-defined on all classes (which can be larger than 2 points, of course). You could use that $x \sim y$ iff there are finitely many steps via generating pairs (or their inverse, of course) from $x$ to $y$.
The unicity is clear, like I said, from having to commute with $\alpha_1$ and $\alpha_2$ $u$ has to be defined like this; it just remains to verify it is actually well-defined.
 Akhil's remark in the comments above merits mention too, as I forgot: we defined $u$ on the disjoint sum originally, showed we had no conflict with the equivalence relation $\sim$, so that we have a well-defined map from the quotient. As the map from the sum is continuous (it is iff both "summands" $\alpha_1$ and $\alpha_2$ are) the standard theorem on quotient spaces (sometimes called the universal theorem for quotient maps) implies that $u$ is indeed continuous from the quotient to $Q$.