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

Matematleta
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2 Answers2

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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.

Kevin Carlson
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  • I attempted a proof using the UMP of the quotient, which I posted as an answer because it seems to be more an answer than a comment. What do you mean by "map out of $X/A$ is continuous...restriction to $X$"? How can we regard $X\subseteq X/A$? (cont.) – Matematleta Aug 25 '18 at 19:33
  • (cont,) I understand the UMP to mean that if a function $f$ out of $X$ preserves $\sim$, then it induces a unique map out of the quotient. This is just a coproduct construction,right? So now I wonder if there is a functor which is a left adjoint, that might be used to provide a quick proof. Something like $-\times X$ (which won't work here). I upvoted your answer, and if you elaborate a bit on it, I will be happy to accept it. Thanks in advance, – Matematleta Aug 25 '18 at 19:33
  • I mean restriction in the generalized sense of precomposing with the canonical map $X\to X/A$. Sorry to be unclear. This is not a coproduct construction, but it is a more general colimit: specifically, it's exactly the pushout you already introduced. The pushout is left adjoint to the functor sending a space $B$ to the diagram $B\leftarrow B\to B$, but this is a direct rephrasing of all the characterizations we've already been discussing. – Kevin Carlson Aug 25 '18 at 21:50
  • I know colim $\dashv \Delta$ but I was trying to say that I showed that $X/*$ is a pushout, but what about the general case of $X/A$? I thought about trying to relate it to a coproduct. We have objects $X$ and $X/\sim $ and a map $\pi:X\to X/A$ and if $f:X\to Z$ is another map that preserves $\sim$, then it factors uniquely through $X/A.$ Now that I think about it, it's a coequalizer , just like in Set, of the diagram $R \rightrightarrows X$ where $R\subseteq X\times X$ s.t. $(x,y)\in R\Leftrightarrow \pi(x)=\pi(y),$ where the arrows are the coordinate projections. Am I in the right track? – Matematleta Aug 26 '18 at 00:33
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    Yes, the quotient by an equivalence relation can always be described by such a coequalizer. – Kevin Carlson Aug 26 '18 at 06:21
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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.

Matematleta
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