Homeomorphisms of topological spaces, are the topologies isomorphic?

Say we have two topological spaces $(X, \tau_X )$, $(Y, \tau_Y )$. As I understand it, a homeomorphism between two topological spaces is an isomorphism $\phi : X\rightarrow Y$, but both $\phi$ and $\phi^{-1}$ have the additional property of being continuous, which means that the pre-image of open sets in the codomain are open i.e. $\forall V\in \tau_Y$ we have $\phi^{-1}(V) \in \tau_X$ and the image of open sets in the domain is open i.e. $\forall U \in \tau_X$ we have $\phi(U)\in \tau_Y$.

As far as I can tell, it is the continuity of $\phi$ which makes it respect the additional structure of topologies. But does the continuity imply that the topologies $\tau_X$ and $\tau_Y$ are isomorphic as sets? Quoting wikipedia:

Homeomorphisms are the isomorphisms in the category of topological spaces—that is, they are the mappings that preserve all the topological properties of a given space.

I mean surely two isomorphic sets with isomorphic topologies are topologically equivalent, right? I'm just wondering if the converse to this statement is true?

Thanks for any replies!

• Homeomorphisms induce a bijection between the two topologies, and are inclusion-preseving. What do you mean by "isomorphic topologies"? Commented Oct 4, 2014 at 21:57
• Oh ok, that's good to know. I mean isomorphism of sets. Commented Oct 4, 2014 at 21:59
• An isomorphism of sets is a bijection. Commented Oct 4, 2014 at 22:14
• Yeah my bad, I do understand that they're the same thing; just my phrasing was poor. Commented Oct 4, 2014 at 22:20
• A topology also has a canonical lattice structure, and a homeomorphism induces an isomorphism of lattices. (Same for $\vee$-complete lattices, bounded lattices.) Commented Aug 29, 2018 at 17:27

to expand on @Crostul's comment:

for a given mapping $f:X \to Y$ between two sets continuity is a relative notion, depending also on the topologies on domain and codomain.

$f$ is continuous iff $f^{-1}(\tau_Y) \subset \tau_X$. here we view a topology on $S$ as a subset of $\mathfrak{P}(S)$ satisfying the usual conditions for open sets.

another way of saying this is that the topology on $X$ pulled back via $f$ from $Y$ is weaker than the original topology on $X$ ('weaker' does not rule out equality).

in the case of bijection $f^{-1}$ is also a mapping hence continuity in this direction requires a similar condition $f(\tau_X) \subset \tau_Y$ i.e. the topology pushed out from $X$ via $f$ to $Y$ is weaker than the original topology on $Y$

since $f$ and $f^{-1}$ are bijective, continuity in both directions forces a bijection between the topologies.

What I think you are referring to is the concept of isomorphism in a category.

• In the category of Topological spaces Top, homeomorphisms are isomorphims of objects in Top which are topological spaces.

• In the category of Sets, Set, bijections are isomorphisms of sets.

• In the category of Groups,Grp, group isomorphisms are isomorphisms of groups.

Isomorphisms sort of give us a loose notion of 'equivalence' in a category.