If $f: X\to Y$ and $g: Y\to X$ are continuous bijections, not necessarily inverses, is $X\cong Y$

Question

I was thinking about to show two sets are bijective, it suffices to construct an injective function both ways. That is, if $f: X\to Y$ and $g: X\to Y$ are both injective (or both surjective), then there exists a bijection between $X,Y$.

But, what if $X, Y$ were topological spaces, and $f, g$ were continuous? Clearly, surjective isn't enough, because I can construct surjective mappings both ways between the interval and the circle, but the two spaces aren't homeomorphic. But, I first notice that this fails because the canonical quotient map from the interval to the circle isn't injective. What if we strengthen this condition to ensure that $f, g$ are continuous AND bijective?

Intuitively, this makes sense: I can continuously map from one space to the other, and my intuition of a homeomorphism is I can bendy-wendy squiggly-wiggly one topological space into the other without rips or gluing together.

But, something tells me this isn't sufficient, because I don't see why $f: X\to Y$ and $g: Y\to X$ continuous bijections guarantee there exists some $h: X\to Y$ such that $h, h^{-1}$ are continuous bijections.

Answer 1

This is based on one example in the MathOverflow question.

Let $X=\mathbb R\sqcup U$ and $Y=(\mathbb R\times\{0,1\})\sqcup U$ where $U$ is a discrete topology on a set with cardinality the continuum, and $\mathbb R$ is the usual space.

Then, we use any bijection $b:U\to U\sqcup \mathbb R.$ Since $U$ is discrete, $b$ is continuous.

Then for $x\in X,$ $$f(x)=\begin{cases}(x,0)&x\in \mathbb R\\(b(x),1)&x\in U, b(x)\in\mathbb R\\b(x)&x\in U, b(x)\in U\end{cases}$$

Also, given any bijection $c:U\to\{0\}\sqcup U,$ then for $y\in Y,$ $$g(y)=\begin{cases}e^w&y=(w,0)\\-e^w&y=(w,1)\\c(y)&y\in U \end{cases}$$

Both of these are continuous bijections.

But $X$ has only one connected component with more than one point, and $Y$ has two.

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Of course, you can write $Y=X\sqcup \mathbb R,$ but it is harder to define $f,g$ from this definition.

Answer 2

From that same thread (I posed the question at the time):

Let $X$ and $Y$ be topological spaces whose underlying sets are $\mathbb{R}$. As topological spaces, $X$ is the disjoint union of the open interval $(0,\infty)$ with a discrete space whose points are nonpositive reals, while $Y$ is the disjoint union of $(-1,0)$, $(1,\infty)$, and a discrete space whose points form the complement of those intervals. Translation by adding one is a continuous bijection from $X$ to $Y$, and also a continuous bijection from $Y$ to $X$, but the two spaces are not homeomorphic (because $X$ has only one connected component that is non-trivial and $Y$ has two).