Suppose we have a continuous function $f : E \to GL_n(\mathbb C)$ where $E$ is an open subset of $\mathbb C$. Can we continuously extend this function to whole space $\mathbb C$ with range in $GL_n(\mathbb C)$.
I think by Tieze's Extension Theorem, we can extend this to a continuous function $\hat{f}: \mathbb C \to \mathcal M(n \times n; \mathbb C)$. But can we do this with the range to be $GL_n(\mathbb C)$?
My idea: can we just make $\hat{f}$ to be constant on $\mathbb C \setminus E$ with value to respect continuity. Not sure this is right.
Just to point out I mis-remember Tieze theorem, the domain must be a closed set. Anyway, there is a great answer below by Eric Wofsey.
No, not in general. For a really simple example, let $n=1$, $E=\mathbb{C}\setminus\{0\}$, and let $f:E\to GL_1(\mathbb{C})=\mathbb{C}\setminus\{0\}$ be the identity map. Then $f$ cannot be extended continuously to $\mathbb{C}$.
More generally, there will be no hope of extending $f$ if there is a sequence approaching a point of $\mathbb{C}\setminus E$ on which $f$ approaches a non-invertible matrix. But even $f$ can extend continuously to $\overline{E}$, there may be homotopy-theoretic obstructions to extending it to $\mathbb{C}$. Specifically, if $f$ extends over $\mathbb{C}$, then $f$ must be nullhomotopic since $\mathbb{C}$ is contractible. For instance, in the example above, if you change $E$ to the annulus $1<|z|<2$, then $f$ extends continuously to $\overline{E}$, but cannot be extended to all of $\mathbb{C}$ because the induced map on $\pi_1$ is nontrivial and so $f$ is not nullhomotopic.