# Why continuous function can be considered as “topological homomorphism”?

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?

• Have you read en.wikipedia.org/wiki/Homeomorphism? – user7530 Nov 26 '16 at 5:22
• 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 5:24
• I think this is a duplicate, though it is an important question. See Qiaochu's answer here: math.stackexchange.com/questions/932942/… – symplectomorphic Nov 26 '16 at 5:36
• @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 6:05
• @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 6:20

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.

• Just as I mentioned in my post, can't injective open map be considered as topological homorphism? – user 170039 Nov 26 '16 at 6:07
• @user170039: I certainly don’t think that it’s a good fit for the concept. – Brian M. Scott Nov 26 '16 at 6:07
• Can you explain the reason? – user 170039 Nov 26 '16 at 6:08
• @user170039: My answer is in large part an explanation. – Brian M. Scott Nov 26 '16 at 6:10
• 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

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.