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In Munkres's book on topology, the notion of homeomorphism is stated to be analogous to the notion of isomorphism in context of modern algebra. I was wondering what will be the analogous concept of homomorphism in context of topology. One of my professors said that it is the continuous functions but I don't understand (although he tried) the reason behind this assertion.

For example in case of group homomorphism we see that a group homomorphism $\varphi:(G,\circ)\to (H,\bullet)$ is a map such that $\varphi(x\circ y)=\varphi(x)\bullet\varphi(y)$ for all $x,y\in G$. If we try to define the notion of, say, "topological homomorphism", in an analogous manner we could define it in the following,

A topological homorphism $\tau:(X,\mathscr{T}_X)\to (Y,\mathscr{T}_Y)$ is a map such that it preserves the "topological structures".

But since here I don't know the precise notion of topological structures, I can't relate the notion of topological homomorphism as stated above to the notion of continuous functions. To me it seems that the notion of injective open map could serve as a notion of "topological homomorphism". Because actually the problem (at least for me) is that while we are discussing groups we can say that the homomorphism is "structure preserving" in the sense that it is "binary operation preserving". But here in case of topological spaces what can play the role of "binary operation"? If we say that the topological homomorphism should preserve the arbitrary union and finite intersection of open sets then the most natural way to think about it is probably the notion of an injective open map.

Can anyone explain this to me?

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    Have you read https://en.wikipedia.org/wiki/Homeomorphism? – user7530 Nov 26 '16 at 05:22
  • You can find a continuous and locally injective function $\mathbb{R} \to S^1$ but there is no continuous and locally injective function $S^1 \to \mathbb{R}$ – reuns Nov 26 '16 at 05:24
  • @user7530: Yes. –  Nov 26 '16 at 05:26
  • @user1952009: Was your comment intended to be part of the explanation that was asked in the question? –  Nov 26 '16 at 05:29
  • It might be better to think at a somewhat higher level in terms of what invariants are to be preserved. Continuous functions preserve compactness, connectedness, and many other important topological invariants (a topological invariant is a property of a topological space preserved under any homeomorphism). Of course they don't preserve all topological invariants, but similarly homomorphisms don't preserve all algebraic properties either. – Ian Nov 26 '16 at 05:29
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    I think this is a duplicate, though it is an important question. See Qiaochu's answer here: http://math.stackexchange.com/questions/932942/why-are-continuous-functions-the-right-morphisms-between-topological-spaces – symplectomorphic Nov 26 '16 at 05:36
  • The structure to be preserved is convergence. The relation of a convergent sequence to its limit can be thought of as a kind of "partially defined infinitary operation", analogous to the algebraic operations in say a group. So a "homomorphism" of topoological spaces should preserve convergence of sequences: $x_n\to x\implies f(x_n)\to f(x).$ For real functions that's enough to characterize continuity (assuming the axiom of choice). In general topological spaces you need some more general notion of convergence, such as nets or filters. – bof Nov 26 '16 at 05:39
  • @user170039 As it happens, the image of a function does not commute with the intersection operation (even just the binary one). The preimage does. This might be relevant. – Ian Nov 26 '16 at 05:55
  • @Ian: That's why I suggested an injective open map to be a possible candidate for the notion of topological homorphism. –  Nov 26 '16 at 05:59
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    @symplectomorphic: No, that’s a significantly different question. Continuous functions are the right category-theoretic choice for morphisms, but in significant other respects they aren’t really the closest analogue of homomorphisms. – Brian M. Scott Nov 26 '16 at 06:05
  • @Brian M. Scott: The questions are only superficially different, so far as I can tell. This one is asking why open maps aren't considered topological homomorphisms. This is implicit in the other post. (Though let me make it clear that I'm not advocating that this question be closed. Perhaps I should simply say the questions are related rather than duplicates.) – symplectomorphic Nov 26 '16 at 06:08
  • @symplectomorphic: I don’t agree. – Brian M. Scott Nov 26 '16 at 06:09
  • @Brian: can you explain why? From your first comment I don't really follow what you think is different about the two questions. I ask only because I suspect I will learn something from hearing what you think the difference is. – symplectomorphic Nov 26 '16 at 06:12
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    @symplectomorphic: On reading the questions more closely, I see that they’re closer than I thought when I read the original version of this one. I am not, however, really satisfied with any of the answers there. – Brian M. Scott Nov 26 '16 at 06:20
  • @Brian M. Scott: I bet many of us would learn something from hearing you argue with Qiaochu's answer and bof's comment (on this post, above) about convergence playing the role of an infinitary algebraic operation to be preserved. – symplectomorphic Nov 26 '16 at 06:22
  • @symplectomorphic: Qiaochu and I have very different approaches to mathematics in general. In particular, he tends to adopt a category-theoretic outlook, something that I as a general and set-theoretic topologist never found at all useful. I suspect that that alone already explains much of the difference in our answers to this question. I’ve commented on bof’s comment in a comment to the OP below my answer. – Brian M. Scott Nov 26 '16 at 06:34

2 Answers2

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Continuous maps aren’t really a very good analogue of homomorphisms in a structural sense; they do preserve some structural features, like compactness and connectedness, but they fail miserably to preserve others. Open maps have the same failing, though they preserve different features. The best analogue is probably the quotient map.

Homomorphisms of algebraic objects correspond naturally to congruence relations on the objects, which in turn correspond to (certain) partitions of the objects. Similarly, quotient maps correspond to partitions of the original space: a surjection $f:X\to Y$ is a quotient map if and only if the open sets in $Y$ are precisely those whose inverse images under $f$ are open in $X$. The map $f$ naturally induces a partition of $X$ into the fibres (point inverses) of $f$. Conversely, if $\mathscr{P}$ is a partition of $X$, the inclusion map $f:X\to\mathscr{P}$ taking $x\in X$ to the unique member of $\mathscr{P}$ containing $x$ becomes a quotient map when $\mathscr{P}$ is given the topology

$$\left\{\mathscr{U}\subseteq\mathscr{P}:f^{-1}[\mathscr{U}]\in\tau(X)\right\}\;,$$

where $\tau(X)$ is the topology of $X$. The resulting quotient space is exactly like $X$ after various chunks of $X$ have been ‘squashed’ to single points, just as a quotient of a group $G$, for instance, preserves all of the structure of $G$ that that is visible after various chunks have been squashed to single elements.

Note too that a continuous bijection is not in general a homeomorphism, while a bijective homomorphism is an isomorphism. A bijective quotient map, however, is a homeomorphism.

Still, thinking of continuous maps as a rather sloppy analogue of algebraic homomorphisms is sometimes helpful to one’s intuition: they do preserve many important topological properties.

Brian M. Scott
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  • Just as I mentioned in my post, can't injective open map be considered as topological homorphism? –  Nov 26 '16 at 06:07
  • @user170039: I certainly don’t think that it’s a good fit for the concept. – Brian M. Scott Nov 26 '16 at 06:07
  • Can you explain the reason? –  Nov 26 '16 at 06:08
  • @user170039: My answer is in large part an explanation. – Brian M. Scott Nov 26 '16 at 06:10
  • If I'm not mistaken then your answer explains the reason for which you think that quotient map is a good choice for the concept of topological homomorphism but it doesn't explain the reason for which you don't think that an open injective good fit for the concept of topological homomorphism. For example to me it seems natural (I admit that "naturality" differs from person to person) to consider an open injective map to be a good fit for the concept of topological homomorphism for the reason I elaborated in my post. –  Nov 26 '16 at 06:19
  • @user170039: Open injective maps simply don’t behave at all like algebraic homomorphisms. Fundamental to the notion of homomorphism is that it allows coarsening of the structure – Brian M. Scott Nov 26 '16 at 06:21
  • @user170039: I’m afraid not: it’s 1:20 a.m. here, and I really need to go fix dinner! I didn’t realize that this answer was going to provoke such a reaction. – Brian M. Scott Nov 26 '16 at 06:23
  • Don't get me wrong. Your answer very naturally corresponds to the notion of topological homorphism for the reason you said. I just want to know what exactly was wrong in my explanation when I said that injective open maps can be considered as topological homorphisms. –  Nov 26 '16 at 06:26
  • @user170039: I think that it really comes down to just what aspects of algebraic homomorphisms one thinks are most essential. To me it’s the association with congruence relations and quotients; quotient maps do that and are structure-preserving in much the same way that homomorphisms are. One can, of course, pick out some structural features that one considers essential and ask what maps preserve them; that’s more of less what bof did above. I agree that his is a very natural notion of structure-preserving; I just don’t think that it matches well with the notion of homomorphism. It’s ... – Brian M. Scott Nov 26 '16 at 06:31
  • ... structure-preserving in a very different (albeit very important) way. (I’ll check back later or tomorrow, but I’m going offline now.) – Brian M. Scott Nov 26 '16 at 06:31
  • I left a comment for you in the chatroom but probably you were not notified. That's why I am asking here again what I asked there. The quotient maps are reminiscent to the so called natural homomorphisms between a group and its factor group, not from arbitrary groups to arbitrary groups, if I understand it correctly. More specifically, the definition of a topological homomorphism that I have in mind should be of the form (or can be shown to be logically equivalent to the form) I highlighted in my post. Do you think such a definition is possible? –  Nov 27 '16 at 03:47
  • @user170039: If $G$ and $H$ are groups, and $h:G\to H$ is a surjective homomorphism, then $H$ is automatically isomorphic to $G/\ker h$. What you’re calling natural homomorphisms are (up to isomorphism) the only surjective homomorphisms. – Brian M. Scott Nov 27 '16 at 03:49
  • Yes. I understood that slightly after I have written my last comment. That's why I removed it. Thank you very much for giving time to clarify my doubts. –  Nov 27 '16 at 04:01
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    Can you give a concrete example as to why you think that "[o]pen injective maps simply don’t behave at all like algebraic homomorphisms."? To me OP's reason doesn't seem to be bad (although I agree that it requires more precise phrasing). – S. Das Nov 28 '16 at 12:56
  • @S.Das: I don’t really know what you’re looking for, since I don’t see any significant analogy in the first place; it’s at best a very superficial one. Open injections don’t have any of the homomorphism-like properties that I’ve mentioned in my answer or my comments on bof’s point, and above all, they simply aren’t important. – Brian M. Scott Nov 28 '16 at 16:37
  • Can you explain what exactly do you mean by "coarsening of the structure"? –  Dec 04 '16 at 13:58
  • @user170039: Losing structural distinctions. Perhaps this answer will help. – Brian M. Scott Dec 04 '16 at 17:44
  • @BrianM.Scott: Thant's an excellent answer. Thanks for suggesting to me the link. –  Dec 05 '16 at 03:25
  • @user170039: You're welcome. – Brian M. Scott Dec 05 '16 at 04:02
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learnmore's answer is probably the most natural one: continuous maps are the ones that preserve topological structure.

Here's a hint that this is indeed the "right" notion: in general, people "know" that there is some sort, or many, actually, form of duality between algebraic and topological structures. Furthermore, in the commutative case (of groups, algebras, etc...), the one-dimensional representations of your algebraic structure are usually enough to completely determine its structure (Pontryagin duality is probably the most popular of such theorems).

Let's consider a certain class of commutative algebraic structures, namely the category $\bf{CC^*Alg}$ of commutative unital) C-algebra (if you don't know what those are, think of matrix algebras over $\mathbb{C}$, and by $*$ I mean either hermitian matrices or conjugate complex numbers, wherever this makes sense). The one-dimensional representations of a commutative C*-algebra $A$ are the same as the $*$-homomorphisms $A\to\mathbb{C}$, just as in the case of (say finite) abelian groups. So, given $A$, take the set of all $*$-homomorphisms $\Omega(A)=*\operatorname{Hom}(A,\mathbb{C})$. It is possible to topologize $\Omega(A)$, in a very natural manner (namely, with the so-called weak-$*$ topology, in such a way that it becomes a topological space.

Moreover, given a $*$-homomorphism $\phi:A\to B$ between unital commutative C*-algebras, there is a canonical map $\phi^*:\Omega(B)\to\Omega(A)$ given by composition: $\phi^*(f)=f\circ\phi$. It just so happens that this is a continuous map between the spectra, in the usual sense, and in fact every continuous map is of this form.

In more precise terms, there is a dual equivalence of categories $\bf{CC^*Alg\to Cpt}$, between the categories of unital commutative C*-algebras and Hausdorff compact spaces, which to a C*-algebra $A$ associates its spectrum, and to a compact Hausdorff space $X$ associates its function algebra $C(X)=C(X,\mathbb{C})$.

All of this means is that everything works just as it should, with the given definition of continuous maps. The most tractable topological spaces are the compact Hausdorff ones, and they have algebraic analogues where continuity translates to usual morphisms of algebraic structures.


I just checked the comments, and Qiaochu Yuan's answer (https://math.stackexchange.com/a/933751/58818) is in a similar spirit to mine's, however overall clearer and more detailed (i.e., better), as I would expect.

Luiz Cordeiro
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