# Unifying Connection Between Topological Embeddings and Quotient Maps

In a book on topology I'm reading the following theorem seemed striking to me, not for its proof, which I believe I understand, but because there's some nice symmetry going on that I'd perhaps like demystified:

Theorem. Let $f:X\to Y$ be a continuous map that is either open or closed.

• If $f$ is injective then it's a topological embedding.
• If $f$ is surjective then it's a quotient map.

If anyone has a good explanation on why these two notions are so closely related, I'd really appreciate it. Thanks!

The duality is that between extremal monomorphisms and extremal epimorphisms. A monomorphism is the right generalization of injective function to situations with more structure, such as topological spaces with continuous functions. We say $f:X\to Y$ is a monomorphism if, whenever $g,h:Z\to X$ are such that $f\circ g=f\circ h$, we can infer $g=h$. Monomorphisms and epimorphisms of topological spaces are just injective and surjective continuous functions, respectively, so aren't too exciting.

But topological spaces have some relatively unusual properties-not shared by algebraic things like groups or vector spaces (or sets-)summed up by the slogan that the first isomorphism theorem does not hold. In particular, injective maps aren't necessarily homeomorphisms onto their image. The latter are clearly interesting, so we look for some kind of abstract characterization of why they're interesting, and the answer I'm offering is that they're extremal. The general definition is that $f:X\to Y$ is an extremal monomorphism if, whenever $f=g\circ e$ for $e:X\to Z,g:Z\to Y$ with $e$ an epimorphism, $e$ is actually an isomorphism (a general word which means bijection when interpreted in sets, homeomorphism when interpreted in topological spaces.)

If you think about it for a minute, you'll find that this definition reads strangely when $X,Y,Z$ are sets and $f,g,e$ arbitrary functions: $f$ is an injection and $e$ a surjection, so that certainly $e$ is a bijection. And indeed all monomorphisms of sets are extremal, but this is not so for topological spaces! It's a good exercise to prove that the extremal monomorphisms where $X,Y,Z$ are topological spaces and $f,g,e$ continuous functions are exactly the embeddings. Use the fact that the identity map from $(X,\tau)$ to $(X,\tau')$ is an epimorphism but not a homeomorphism whenever $\tau'$ is strictly coarser than $\tau$.

It's a general principle that whenever you have a notion defined entirely in terms of arrows, like monomorphism or extremal monomorphism, you have a dual property gotten by turning all the arrows around: this is the basic duality of category theory. When we turn the arrows around in the definition of monomorphism, we get the notion of epimorphism, and when we do the same for extremal monomorphisms, we get the notion of extremal epimorphism: $f:X\to Y$ is an extremal epimorphism if whenever $f=g\circ e$ and $g$ is a monomorphism, then $g$ is an isomorphism. And in topological spaces, this notion is exactly that of quotient map!

So the upshot is that we like injections and surjections, and have various generalizations of them, many of which end up being interesting in any new context we begin to study. And there's always a duality between such generalizations extending the duality between monomorphisms and epimorphisms. So as soon as you notice embeddings are interesting, you can immediately begin searching for quotient maps! Of course, that's a ludicrously ahistorical description, but it's conceptually not entirely useless.

For completeness' sake, let me mention that I could have also told this whole story with regular monomorphisms and epimorphisms, which are the analogue of normal subgroups (and quotient groups) from group theory. In general these are much stronger than extremal mono/epis, but in the case of spaces they coincide.

• While interesting, I don't see how this answers the OP's question. So they're dual; so what? This doesn't explain why if $f$ is a closed mapping, then injective implies embedding and surjective implies quotient, nor does it explain why if $f$ is an open mapping, then the same thing happens. – goblin GONE May 26 '15 at 3:58
• @goblin I took the question to be "why are quotients and embeddings so closely related," and the OP apparently disagrees with you about whether this answers the OP's question. – Kevin Arlin May 26 '15 at 4:01
• Look, there's clearly something more subtle than mere categorical duality at work here. If it was mere duality the OP was observing, they would have observed: "property $P$ + injective implies embedding" and "dual of property $P$ + surjective implies quotient." What he has actually observed is something more like "property $P$ + injective implies embedding, property $P$ + surjective implies quotient." I find this genuinely remarkable. I'm not saying your answer is wrong or anything, and I certainly wouldn't want it deleted, but there's more to this question than you're giving it credit for. – goblin GONE May 26 '15 at 8:00
• I think it's a good answer. It explains why quotients and embeddings are related, which the poster certainly seems curious about. – ಠ_ಠ May 26 '15 at 11:17
• @goblin I did appreciate this answer but if you have more to add I'd certainly appreciate another answer from you – Samuel Yusim May 26 '15 at 15:01