Two manifolds $X$ and $Y$ with two bijective continuous functions from $X$ to $Y$ and $Y$ to $X$, but not homeomorphic I happen to found this question about finding two metric spaces $X$ and $Y$ such that there exists bijective continuous functions $f:X\to Y$ and $g:Y\to X$, and $X$ and $Y$ are not homeomorphic. I could not produce my own example, and I think the reason is that I am thinking of manifolds only.
Suppose we have two second-countable Hausdorff metrisable manifolds $X$ and $Y$ such that there exists bijective continuous functions $f:X\to Y$ and $g:Y\to X$. Can we show that the two manifolds are homeomorphic? Or is there still a counterexample?
Please note that I have taken only one course on metric space and topology, and I don't have much experience with manifolds. So please explain answers in details or post references to the details.
Edit: In case of any ambiguity, I would like to state that I do not restrict attention to manifolds without boundary, i.e. the manifolds may have boundary, if such a counterexample exists.
 A: To understand this answer you should know about how one can obtain a torus or a cylinder via a projection from the unit square. 
To do this, one simply identifies opposite sides of the square: 2 for the cylider, and 4 for the torus. On Wikipedia there is a nice gif which illustrates this.
Now for the answer. Denote by $R$ the half open unit rectangle, by $C$ the corresponding half open cylinder, and by $T$ the torus. One has projecitons 
$$
\pi_{R, C} : R \to C, \quad \pi_{C, T} : C \to T, \quad \pi_{R, T} = \pi_{C,T} \circ \pi_{R, C} : R \to T
$$
The spaces we need, are the following 
$$
X = \amalg_{\mathbb{N}}R ~\cup~ C ~\cup~ \amalg_{\mathbb{N}} T\\
Y= \amalg_{\mathbb{N}}R ~\cup R ~\cup~ \amalg_{\mathbb{N}} T
$$
The maps are the following:
From $X$ to $Y$, you project to the right, i.e. $C$ maps to $T$, and then you continue, by mapping the remaining $T$ components to themselves, and the $R$ to themselves as well:
$$
X = \amalg_{\mathbb{N}}R ~\cup~ C ~\cup~ \amalg_{\mathbb{N}} T\\
~~\downarrow~~~~ \searrow \quad \searrow ~~~\downarrow\\
Y= \amalg_{\mathbb{N}}R ~\cup R ~\cup~ \amalg_{\mathbb{N}} T
$$
From $Y$ to $X$ you just project upward, i.e. the middle $R$ projects to the single $C$, and the rest is just identities.
$$
X = \amalg_{\mathbb{N}}R ~\cup~ C ~\cup~ \amalg_{\mathbb{N}} T\\
~~\uparrow~~~~~~ \uparrow ~~~~~~~~~\uparrow\\
Y= \amalg_{\mathbb{N}}R ~\cup R ~\cup~ \amalg_{\mathbb{N}} T
$$
Since $X$ and $Y$ are not homeomorphic (left as an exercise), this would be a counterexample.
