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This question comes from Conlon's Differentiable Manifolds (it's Exercise 1.1.13).

Let $X$ and $Y$ be connected, locally Euclidean spaces of the same dimension. If $f:X \rightarrow Y$ is bijective and continuous, prove that $f$ is a homeomorphism.

I think I need to use the local homeomorphism $\Rightarrow$ global homeomorphism idea, but I'm having trouble constructing the local homeomorphism. Obviously if I were to do so these local homeomorphisms would have to come from composing the homeos that we have into $\mathbb{R}^n$, but I keep having trouble because we don't know $f$ is a homeo yet. Am I even on the right track?

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I won't claim this is a full answer, but I think you may want to show that any bijective continuous function between to open sets in euclidean space is an open map, and then this should follow. EDIT: This may be using a flamethrower to get rid of leaves in the lawn,… – John Stalfos Apr 13 '12 at 13:51
@JohnStalfos A bijective continuous function between two open sets in a Euclidean space should be a local homeomorphism because a bijective continuous function from a compact space to a Hausdorff space is a homemorphism. Now a bijective local homeomorphism is a homeomorphism. Doesn't this settle what you state without Invariance of Domain? – caffeinemachine Jun 4 '15 at 20:15

OK, try this.

It is enough to show that $f^{-1}$ is continuous. So suppose $f(x) = y$ and $x \in U_x$ where $U_x$ is open. We want an open $U_y$ such that $y \in U_y \subseteq f(U_x)$.

Since $Y$ is locally Euclidean, we can get an open $B_y \simeq \mathbb{R}^n$ with $y \in B_y$. $f$ is continuous, so $f^{-1}(B_y)$ is open. $U_x \cap f^{-1}(B_y)$ is also open. Since $X$ is locally Euclidean, we can get an open $B_x \simeq \mathbb{R}^n$ with $x \in B_x \subseteq U_x \cap f^{-1}(B_y)$.

Now consider the map $f': \mathbb{R}^n \simeq B_x \to f(B_x) \to B_y \simeq \mathbb{R}^n$. Each step is injective and continuous, so $f'$ is also injective and continuous. Now we can use Invariance of Domain (thank you @John Stalfos!) to conclude that $f'$ is open. In particular, $f(B_x)$ is open in $B_y$, and since $B_y$ is open in $Y$, $f(B_x)$ is open in $Y$. Now it's clear that we can choose $U_y$ above to be $f(B_x)$. QED.

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