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I have a practice exam problem that asks to prove or disprove the following statement.

Every continuous bijection between homeomorphic spaces is a homeomorphism.

Now based on everything I know about topology, I feel like I have several ways to show that this is indeed false. But then I started second guessing myself based on the wording of the question. It's kind of hard for me to interpret what this question is really asking.

So we already know that we have homeomorphic spaces $(X,\tau_1), (X,\tau_2)$. But are they really homeomorphic if $\tau_1\neq \tau_2$?

I guess what I'm really confused about is what is meant by homeomorphic spaces precisely. Because I know that a continuous bijection is a homeomorphism if and only if it is open (closed), which is equivalent to saying that a continuous bijection is a homeomorphism if and only if $\tau_1=\tau_2$. But if they had to be equal in order for the spaces to even be homeomorphic to begin with, then can't we argue that any continuous bijection between the spaces has to be a homeomorphism?

Peter Mortensen
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ai.jennetta
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2 Answers2

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The simplest sort of example to keep in mind is the following: let $X$ be a set with more than two elements, and endow $X$ with the discrete topology (in which every subset of $X$ is open). Then every function with domain $X$ is continuous. Now write $Y$ for the topological space whose underlying set is $X$, but with any topology that is strictly coarser than that of $X$ (for instance, the trivial topology in which the only open subsets are the empty set and $X$). Then the identity map $X \rightarrow Y$ is a continuous bijection which is not a homeomorphism.

To your question, the two different topologies on $X$ are the same if and only if the identity map is a homeomorphism between them; more generally, given a bijection $\phi: X \longrightarrow X$, the map $\phi$ is a homeomorphism if and only if $\tau_2=\phi(\tau_1)$.

For an example of a space with a continuous bijection to itself that is not a homeomorphism, take $X=\mathbf{Q}_{>0}$ to be the positive rational numbers with the topology consisting of sets of the form $(m,\infty)$, for $m \in \mathbf{Z}_{\geq 0}$, together with $\emptyset$. Take $\phi$ to be the multiplication by $1/2$ map. Evidently $\phi$ is a continuous bijection, but it is not a homeomorphism.

Stephen
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Because I know that a continuous bijection is a homeomorphism if and only if it is open (closed), which is equivalent to saying that a continuous bijection is a homeomorphism if and only if $\tau_1=\tau_2$.

This is not true. For example take $\Bbb R$ with the topologies

\begin{align} \tau_1 &= \{(a,b)\mid a<0<b\},\\ \tau_2 &= \{(a,b)\mid a<1<b\}. \end{align}

The topologies are different . However, these spaces are homeomorphic because there is the map $\varphi:x\mapsto x+1$ which is a homeomorphism. It is not even neccessary that the two topological spaces have to be defined on the same base space. A topological space $(\Bbb R,\tau_1)$ can be homeomorph to a space $(\Bbb R^2,\tau_2)$ for appropriately defined topologies.

Being homeomorphic only means that the topologies are essentially the same with respect to anything that is important in the subject of topology. But not that they are the same from a set theoretic point of view.

Every continuous bijection between homeomorphic spaces is a homeomorphism.

I am not sure if your question contains that you ask us to answer this question, but just in case, here is a counter example:

Take $X=\Bbb N\times[0,1)$ with the product topology (subspace topology on $\Bbb N$ and $[0,1)$ respectively). You can always take two of the intervals $[0,1)$ and join them into a single one by gluing their bounds. This is continuous, bijective and leaves behind the same space $\Bbb N\times [0,1)$. But it is not a homeomorphism because the inverse is not continuous (because of cutting a single interval into two).

Further, this question has very nice answers of the same task, where $X$ is even connected.

M. Winter
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