Why does invariance of domain theorem apply to open sets homeomorphic to Rn?

[Rasalhague]

In the first volume of Differential Geometry, Ch. 1, Spivak states that if $U \subset \mathbb{R}^n$ is homeomorphic to $\mathbb{R}^n$, then $U$ is open. This seems obvious: $\mathbb{R}^n$ is open in $\mathbb{R}^n$, so its pre-image under a homeomorphism $f:U \rightarrow \mathbb{R}^n$ is open. The pre-image under $f$ of $\mathbb{R}^n$ is $U$. Therefore $U$ is open in $\mathbb{R}^n$.

Why does Spivak not take this obvious route? Am I mistaken about it? Instead, he says that proof of the openness of $U$ needs something called the Invariance of domain theorem.

 

[Hurkyl]

The pre-image under $f$ of $\mathbb{R}^n$ is $U$. Therefore $U$ is open in $\mathbb{R}^n$.

This step is wrong. The correct conclusion from that information is

... therefore U is open in U.​

 

[Rasalhague]

Ah, I see! And, of course, U must be open in U anyway, being a topological space; but the issue is whether U is open in $\mathbb{R}^n$. Thanks, Hurkyl.

 

[Rasalhague]

I still don't get it. The IOD theorem is of the form "(A&B) implies C". Since f is a homeomorphism, B and C are true (f is 1-1 and continuous, and f is a homeomorphism). But this says nothing about A (U is open in R^n). (A&B) implies C" is consistent with A being true or false.

As an aside, does the theorem assume that f is onto and hence invertible, or is this implied by the antecedents?

 

[alexfloo]

Replace C with C&D. I believe you're correct that D is trivial, and it was strange to include it. However, C is still certainly nontrivial (the openness of V in ℝⁿ).