Here is Rotman's proof for the theorem:

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(Theorem 6.31 says that if $S^n$ contains an $r$-cell $e_r$, then $S^n-e_r$ is acyclic. Theorem 6.35 is the Jordan-Brouwer Separation Theorem.)

However, the "Invariance of Domain" theorem is often stated as following:

Theorem 1. If $U$ is an open subset of $\mathbb{R}^n$ and $f : U \rightarrow \mathbb{R}^n$ is an injective continuous map, then $V = f(U)$ is open and $f$ is a homeomorphism between $U$ and $V$.

I tried to generalize the theorem on Rotman's book get the above result. Then I started to realize that the condition of "homeomorphism" might be unnecessary: The proof still works if $h$ is only bijective and continuous. Is that right? If it's right we can show $f$ is an open map in Theorem 1. The result then follows. Are there any mistakes I made?


Yes, this is correct. The key point is that the proof only actually uses the restriction of $h$ to $N$, which is then automatically a homeomorphism to its image since $N$ is compact.

Alternatively, you can use a similar idea to deduce the statement for continuous injections from the statement for homeomorphisms. For if $U\subseteq\mathbb{R}^n$ and $f:U\to\mathbb{R}^n$ is a continuous injection, consider any open ball $B$ such that $\overline{B}\subseteq U$. Then $f$ is a homeomorphism to its image when restricted to $\overline{B}$, and thus a homeomorphism to its image when restricted to $B$. The result for homeomorphisms now gives that $f(B)$ is open in $\mathbb{R}^n$. Since every open subset of $U$ is a union of such balls $B$, this shows $f$ is an open map, and hence a homeomorphism to its image.


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