The determinant is an open map

Idea. For an open $U\ni A$ we must find an interval around $\det A$ contained in $\det(U)$. We move $A$ along a curve inside $U$ along which $\det$ varies over an interval around $\det A$.

Case 1: $\det A\ne0$. Let $A_s=A\operatorname{diag}(s,1,\dots,1)$. Then $\det A_s=s\det A$ and $A_s\to A$ as $s\to1$, so $A_s\in U$ for $s\in(1-\tau,1+\tau)$ for some $\tau>0$. As $s$ runs over $(1-\tau,1+\tau)$, $\det A_s$ runs over the interval $\big(\det A(1-\tau),\det A(1+\tau)\big)$ (or reversed), which is a neighbourhood of $\det A$.

Case 2: $\det A=0$. Write the SVD $A=P\Sigma Q^T$ with $P,Q$ orthogonal and $\Sigma=\operatorname{diag}(\sigma_1,\dots,\sigma_r,0,\dots,0)$, $\sigma_i>0$, $r<n$. Let $k=n-r\ge1$. For $\tau>0$ small and $s\in[-\tau,\tau]$ put
$$\Sigma_s=\operatorname{diag}(\sigma_1,\dots,\sigma_r,\,s,\,\tau,\dots,\tau)\quad(\text{with }k-1\text{ entries equal to }\tau),$$
and $A_s=P\Sigma_sQ^T$. Then
- $\|A_s-A\|=\|\Sigma_s-\Sigma\|\le\tau$ (operator norm is invariant under orthogonal maps), so $A_s\in U$ when $\tau$ is small enough;
- $\det A_s=\det P\det Q\cdot\sigma_1\cdots\sigma_r\cdot\tau^{k-1}\cdot s=c\,s$ where $c=\pm\sigma_1\cdots\sigma_r\tau^{k-1}\ne0$.

As $s$ ranges over $[-\tau,\tau]$, $\det A_s=cs$ covers the interval $[-|c|\tau,|c|\tau]$, a neighbourhood of $0=\det A$.

Conclusion. For every open $U$ and $A\in U$, $\det(U)$ contains a neighbourhood of $\det A$, so $\det(U)$ is open.

Remark. The determinant is a *submersion* at every matrix of rank $\ge n-1$ (its derivative $\operatorname{tr}(\operatorname{adj}(A)\,\cdot)$ is nonzero there, Problem 79), so openness is easy near them; the interest is at matrices of rank $\le n-2$ where $d\det=0$: $\det$ is still open though not a submersion. For $n=1$ it is the identity.