Preimage of open pathwise connected set is pathwise connected

Question

Is it a fact that under a continuous function, the preimage of an open, pathwise connected set is pathwise connected itself? I'm trying to prove that $GL_n(\mathbb{C})$ is pathwise connected, without explicitly constructing paths.

Answer 1

No, the preimage of a (pathwise) connected set under a continuous map has no reason to be connected. For examples, consider the projection $\pi \colon X\times Y \to X$ where $Y$ is a discrete space (with more than one point), or maps with discrete domain (these are automatically continuous).

I'm trying to prove that $GL_n(ℂ)$ is pathwise connected, without explicitly constructing paths.

Depending on what counts as explicit, we can prove it more or less easily.

First way, almost explicit paths:

Let $A \in GL_n(\mathbb{C})$. Consider $\varphi \colon \mathbb{C} \to M_n(\mathbb{C});\; \varphi(z) = (1-z)\cdot A + z\cdot I$. Then $p = \det \circ \varphi$ is a polynomial (of degree $\leqslant n^2$) with $p(0) \neq 0 \neq p(1)$, hence has only finitely many zeros. Choose a path $\gamma$ in $\mathbb{C}$ from $0$ to $1$ that avoids the zeros of $p$.

Then $\varphi \circ \gamma$ is a path in $GL_n(\mathbb{C})$ connecting $A$ and $I$.

That was easy.

Second way, no paths, but heavy machinery:

Definition: Let $D \subset \mathbb{C}^n$ open. A subset $E \subset D$ is thin in $D$ if it is locally contained in the zero set of a non-constant holomorphic function, i.e.

$$\bigl(\forall x\in D\bigr) \bigl(\exists U \in \mathcal{V}(x)\bigr) \bigl(\exists f \in \mathcal{O}(U)\setminus\{0\}\bigr)\bigl(E\cap U \subset Z(f)\bigr).$$

Proposition: Let $D \subset \mathbb{C}^n$ open and connected. Let $E \subset D$ thin in $D$. Then $D \setminus E$ is connected.

The proposition is proved in most (hopefully) books on several complex variables, for example, it's corollary 3.6 in chapter I of Range's "Holomorphic Functions and Integral Representations in Several Complex Variables".

Armed with that heavy machinery, we now observe that $E = M_n(\mathbb{C}) \setminus GL_n(\mathbb{C})$ is the zero set of a non-constant holomorphic function (the determinant), hence thin. Therefore $GL_n(\mathbb{C}) = M_n(\mathbb{C}) \setminus E$ is connected, since the vector space $M_n(\mathbb{C})$ is connected. $GL_n(\mathbb{C})$ is also open, hence it is pathwise connected.