Adjunction space is homeomorphic to quotient: an easier proof?

Question

All spaces are in $Top.$ It is trivially obvious geometrically that if $A\subseteq X$, and $f:A\to *$, where $*$ is a singleton, then $X/A\cong X\cup_f *.$

I have never seen a proof of this, however, and since I know some category theory, I did it using pushouts. My work follows. But it seems like a lot of work to show something so trivial so my question is, how could we show this without using category theory?

Consider the following commutative square. I show it is a pushout, from which we may conclude immediately that $X/A\cong X\cup_f *.$

$\require{AMScd} \begin{CD} A @>{f}>> *\\ @V{i}VV @VV{g}V\\ X @>>{\pi}> X/A\end{CD}$

Suppose there are $Z\in Top$ and $\alpha: X\to Z;\ \beta:*\to Z$ such that $\beta f=\alpha i\ .$

Then, of course, $\beta f=z_p\in Z.$ Now, if we define $\phi:X/A\to Z$ by: $\phi([x])=\alpha (x)$ if $x\notin A$ and $\phi([A])=z_p,$ then,

$\phi \pi=\alpha:$

$x\notin A\Rightarrow \phi ([x])=\alpha (x)\ $ and $x\in A\Rightarrow \phi \pi(x)=\phi([A])=z_p\ =\beta f(x)=\alpha i(x)=\alpha (x).$

$\phi g=\beta:$

$\phi g(*)=\phi([A])=z_p=\beta (*)$

The definition of the quotient topology implies that $\phi $ is continuous because $\alpha$ is and $\phi \pi=\alpha\Rightarrow \alpha^{-1}=\pi^{-1}\phi^{-1}.$

$\phi$ is unique by construction.

Therefore, $X/A\cong X\cup_f *.$

Answer 1

I would just say $X/A$ is defined to have the topology such that a map out of $X/A$ is continuous if and only if its restriction to $X$ is, which is the same as the topology on the pushout. This amounts to using more seriously the fact that your two spaces are obviously in continuous bijection, so that we only have to compare the topologies. This isn't exactly how the quotient topology is normally defined, but the equivalence is certainly no more complicated than the proof you gave, and is much more general. Alternatively and preferably, one should define $X/A$ using the universal property in the first place, and compute the topology in proving that such a space exists.

Answer 2

Here is a proof that uses only the UMP of the quotient: Consider the commutative square

$\require{AMScd} \begin{CD} X @>{i}>> X+*\\ @V{p}VV @VV{\pi}V\\ X/A @>>{\phi}> X\cup_f *\end{CD}$

where $\phi$ is the unique (continuous) map induced by $\pi i.\ \phi $ is evidently bijective.

If $C$ is closed in $X/A$ and $[A]\notin C$, then $\pi^{-1}\phi (C)=ip^{-1}(C)=(C,1)$ which is closed in $X+*\Rightarrow \phi (C)$ is closed in $X\cup_f *$

On the other hand, if $[A]\in C$ then $\pi^{-1}\phi (C)=(C,1)\cup (*,2)$ and this set is also closed in $X+*\Rightarrow \phi (C)$ is closed in $X\cup_f *.$

So, $\phi$ is a closed map and therefore a homeomorphism.