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In short: I find it intuitive that isomorphic algebraic structures (be it groups, rings, fields, vector spaces,...) have the same algebraic properties. Yet I do not find it as intuitive that two homeomorphic topological spaces have the same properties (insofar as these properties deal exclusively with their topology). Instead I find it intuitive to define a 'topological isomorphism' not as a homeomorphism, but as I have defined it below so as to follow the pattern of defining isomorphisms in algebra.

The next section can be skipped; it just serves to explain why I find algebraic isomorphisms intuitive, as opposed to homeomorphisms, as well as to why I think homeomorphisms 'break the pattern' in terms of how isomorphisms are defined in algebra. My concise questions are at the end of the post.


As I see it, an group isomorphism $\phi:(G_1,+)\to (G_2,\times)$ is a 'renaming' of the elements of $G_1$ and of its operator: any $x\in G_1$ is renamed to $\phi(x)\in G_2$, and $+$ is renamed to $\times$. Under that view "$x + y = z$" says the same as "$\phi(x)\times\phi(y) = \phi(z)$", yet with different symbols.

Given two isomorphic groups $(G_1, +)$ and $(G_2, \times)$ it is intuitive to me that any 'group-theoretic' property holds in one group if and only if it holds in the other.

A vector space consists of an abelian group of vectors $(V,+_V)$, a field of scalars $(F,+,\times)$, and scalar multiplication $(\cdot):F\times V$. Thus a vector space can be seen as a $6$-tuple $(V,+_V, F, +, \times, \cdot)$. I would define a 'vector space isomorphism' between $(V_1,+_{V_1}, F_1, +_1, \times_1, \cdot_1)$ and $(V_2,+_{V_2}, F_2, +_2, \times_2, \cdot_2)$ as a pair $(f,g)$ of bijections $f:V_1\to V_2$ and $g:F_1\to F_2$ that 'preserve vector space operations' i.e.

  1. $f(v+_{V_1}u)=f(v)+_{V_2}f(u)$ for any $v,u\in V_1$.
  2. $g(a+_1b)=g(a)+_1g(b)$ for any $a,b\in F_1$.
  3. $g(a\times_1b)=g(a)+_1g(b)$ for any $a,b\in F_1$.
  4. $f(a\cdot_1v)=g(a)\cdot_2 f(v)$ for any $a\in F_1, v\in V_1$.

Essentially I'm saying that $f$ is a bijective linear map, that $g$ is a field isomorphism, and that $f$ and $g$ 'preserve' scalar multiplication i.e. $4)$ holds. A linear map is a specific case of the above. As with groups, it is possible to see $f$ and $g$ as 'renaming' the vectors, scalars, and operations between them.

Given two vector spaces and a bijective linear map between them, it is intuitive to me that any 'linear-algebra-theoretic' property holds in one vector space if and only if it holds in the other.

That intuition remains with me for all algebraic structures I have studied, however given two topological spaces and a homeomorphism (often called a "topological isomorphism") between them, it is not intuitive to me that any property (that deals exclusively with their topology) holds in one space if and only if it holds in the other. The reason being that a homeomorphism does not seek to preserve any operators (unlike group isomorphisms, linear maps, etc). If it was defined so as to preserve 'topological operators' it would -I think- be defined as follows:


Definition: given two topological spaces $(X,\tau_X)$ and $(Y,\tau_Y)$ a topological isomorphism $(f,\phi)$ is a pair of bijections $f:X\to Y$ and $\phi:\tau_X\to\tau_Y$ that preserve 'topological operations' i.e.

  1. Given $x\in X$ and $A\in \tau_X$, we have that $x\in A$ if and only if $f(x)\in \phi(X)$.
  2. Given an arbitrary collection $\{A_\alpha\}\subseteq \tau_X$ we have $$\phi\left(\bigcup_\alpha A_\alpha\right) = \bigcup_\alpha \phi(A_\alpha).$$
  3. Given $A,B\in\tau_X$ we have $$\phi\left(A\cap B\right) = \phi(A)\cap \phi(B).$$

The following results are almost immediate. Notably, since $1)$ and $2)$ below deal with concepts which definition does not require us to use "$\in$" (closed sets, basis, countability) their proof requires only the use of $\phi$. However, since the definition of a Hausdorff space uses the $\in$ operator, the proof of $3)$ below does make use of $f$, the function which "translates" $\in$ from $(X,\tau_X)$ to $(Y,\tau_Y$)

Theorem: if there is some topological isomorphism $(f,\phi)$ between $(X,\tau_X)$ and $(Y,\tau_Y)$, then

  1. $A$ is closed in $\tau_X$ if and only if $f(A^c)$ is closed in $\tau_Y$.
  2. $\tau_X$ is second-countable if and only if so is $\tau_Y$.
  3. $\tau_X$ is Hausdorff if and only if so is $\tau_Y$.
  4. $\ldots$

Why are topological isomorphisms (as I have defined them) not ever used in topology?

How do topological isomorphisms relate to homeomorphisms?


PS: I know very little universal algebra or category theory.

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  • $\begingroup$ By definition, a topological property is a property that is preserved by homeomorphisms. $\endgroup$
    – wormram
    Commented Jul 12, 2023 at 22:32
  • $\begingroup$ @morrowmh right, I should change my terminology. $\endgroup$
    – Sam
    Commented Jul 12, 2023 at 22:36
  • $\begingroup$ What you've defined would be related to the topic of point-free topology, where you forget $X$ and $Y$ entirely and instead consider only the algebraic objects $\tau_X$ and $\tau_Y$ (though in that subject, a "morphism of frames" $f \in \operatorname{Hom}(\tau_X, \tau_Y)$ is a function map in the opposite direction which preserves sups and finite meets, in order to correspond to what a continuous function $X \to Y$ induces on the collections of open sets). $\endgroup$ Commented Jul 12, 2023 at 23:16
  • $\begingroup$ Groups and topological spaces have a certain differences and certain commonalities. A group structure on $G$ is built from Cartesian powers. The group operation is a binary operation, which is a certain subset of $G^3=G \times G \times G$. The inverse operation on $G$ is a function from $G$ to itself, which is a certain subset of $G^2=G \times G$. But a topological structure on $X$ is built on power sets: the topology on $X$, which is a certain collection of subsets, is a certain subset of $\mathcal P(X)$, which is a certain element of $\mathcal P(\mathcal P(X))$. $\endgroup$
    – Lee Mosher
    Commented Jul 12, 2023 at 23:53
  • $\begingroup$ So much for the differences. The commonality is: a bijection $f : A \to B$ induces a bijection $A^2 \to B^2$ and a bijection $A^3$ to $B^3$ and a bijection $\mathcal P(A) \to \mathcal P(B)$ and a bijection $\mathcal P(\mathcal P(A)) \to \mathcal P(\mathcal P(B))$. So, for both groups and homeomorphisms, an isomorphism is a bijection that takes the structure to the structure under the appropriate induced bijection. $\endgroup$
    – Lee Mosher
    Commented Jul 12, 2023 at 23:55

1 Answer 1

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Let $\mathcal{X}=(X,\tau), \mathcal{Y}=(Y,\sigma)$. Your topological isomorphisms are equivalent to homeomorphisms in the following sense:

  1. If $f:\mathcal{X}\rightarrow\mathcal{Y}$ is a homeomorphism then there is exactly one function $g:\tau\rightarrow\sigma$ such that $(f,g)$ is a topological isomorphism.

  2. If $(f,g)$ is a topological isomorphism from $\mathcal{X}$ to $\mathcal{Y}$, then $f$ is a homeomorphism from $\mathcal{X}$ to $\mathcal{Y}$.

For $1$, the unique $g$ is the function $U\mapsto\{f(u): u\in U\}$. In fact, note that the "set-map part" always has this form in any topological isomorphism. For $2$, continuity of $f$ and $f^{-1}$ follow from the "$x\in A\iff f(x)\in g(A)$" condition.

I've kept things deliberately sketchy here; writing out the details of this proof is a good exercise, and I think will help build intuition for homeomorphisms. As to why we don't use this notion, I think it's just a matter of efficiency: no information is lost by discarding the "sets-map" part.

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  • $\begingroup$ As it's past midnight here I will do the exercise tomorrow, but I already find your answer helpful so +1. Do you happen to know why the definition I gave is never used? $\endgroup$
    – Sam
    Commented Jul 12, 2023 at 22:54
  • $\begingroup$ @Sam See the last sentence of my answer. $\endgroup$ Commented Jul 12, 2023 at 22:56
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    $\begingroup$ @Sam: Let me weigh in on the question in your latest comment, with a slight spin on the last sentence of this answer: the ordered pair $(f,\phi)$ determines and is determined by its first entry $f$, so it is simpler to let $f$ alone be the "isomorphism" object. $\endgroup$
    – Lee Mosher
    Commented Jul 12, 2023 at 23:00
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    $\begingroup$ The mathematical culture generally prefers definitions (like axioms) to be as spare and efficient as possible. $\endgroup$
    – Lee Mosher
    Commented Jul 12, 2023 at 23:01

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