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.
- $f(v+_{V_1}u)=f(v)+_{V_2}f(u)$ for any $v,u\in V_1$.
- $g(a+_1b)=g(a)+_1g(b)$ for any $a,b\in F_1$.
- $g(a\times_1b)=g(a)+_1g(b)$ for any $a,b\in F_1$.
- $f(a\cdot_1v)=g(a)\cdot_2 f(v)$ 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.
- Given $x\in X$ and $A\in \tau_X$, we have that $x\in A$ if and only if $f(x)\in \phi(X)$.
- Given an arbitrary collection $\{A_\alpha\}\subseteq \tau_X$ we have $$\phi\left(\bigcup_\alpha A_\alpha\right) = \bigcup_\alpha \phi(A_\alpha).$$
- 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
- $A$ is closed in $\tau_X$ if and only if $f(A^c)$ is closed in $\tau_Y$.
- $\tau_X$ is second-countable if and only if so is $\tau_Y$.
- $\tau_X$ is Hausdorff if and only if so is $\tau_Y$.
- $\ldots$
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.
Let $\mathcal{X}=(X,\tau), \mathcal{Y}=(Y,\sigma)$. Your topological isomorphisms are equivalent to homeomorphisms in the following sense:
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.
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.