Suppose that $U$ is an open subset of $\mathbb{R}^{n}$ and $f: \mathbb{R}^{n} \rightarrow \mathbb{R}^{m}$ (for some integers $n$ and $m$) is a continuous bijective function whose Jacobian has nonzero determinant on $U$. We can use the inverse function theorem to show that $f$ has an inverse, $f^{-1}$ that exists and is continuous on $f(U)$. Let $S$ be a subset of $\bar{U} - U$, where $\bar{U}$ is the closure of $U$, i.e. $S$ consists of limit points of $U$. Suppose that $f$ extends to a function $F$ that is continuous and bijective on $V = U \cup S$. Can we prove that $f$ has an inverse $f^{-1}$ that is continuous on $f(V)$? If so, how? It seems like it should be a somewhat trivial consequence of $f^{-1}$ existing on $f(V)$ and being continuous on $f(U)$, but I am not exactly sure how to fill in the details. Any advice would be helpful!
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1Why specify that $n, m$ are integers? What else would they be? – Qiaochu Yuan Aug 02 '24 at 05:31
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This can only work for $n = m$. – Paul Frost Aug 02 '24 at 08:05
1 Answers
Firstly, I assume that $n=m$ so that $f \colon \mathbb{R}^n \to \mathbb{R}^n$, otherwise the determinant isn't defined for its Jacobian.
Additionally, we can just denote $\bar{U} = U \cup S = V$.
You're on the right tract with the idea that this should follow from the inverse of a continuous function being continuous. However, this is only true if $\bar{U}$ is compact (i.e., bounded in $\mathbb{R}^n$).
To see this, note that if $\bar{U}$ is compact then $f(\bar{U})$ is also compact since the continuous image of a compact set is compact. Let $W$ be a closed subset of $\bar{U}$. A closed subset of a compact set is compact, so $W$ is compact $\Rightarrow f(W)$ is compact (and hence closed). Observe, $$ (f^{-1})^{-1}(W) = f(W)$$ so that $f^{-1}$ is continuous (a function is continuous if and only if the preimage of any closed set is closed).
If $\bar{U}$ is not compact (i.e., unbounded) then there are plenty of examples where the inverse fails to be continuous. See for example: this answer.
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