It is well known that the set of $A\in\mathbb{R}^{n\times n}$ such that $\det(A)\ne 0$ is dense in $\mathbb{R}^{n\times n}$. This is one way to show this:
Let $B\in\mathbb{R}^{n\times n}$ with singular values $s_1,...,s_n$. Then for a sequence $\lambda_n$ in $\mathbb{R}$ with $\lambda_n\notin\{s_1,...,s_n\}$ for all $\lambda_n$ and $\lambda_n\to 0$, $\det(B+\lambda_nI)\ne 0$ and $B+\lambda_nI \to B$.
I am wondering if the previous statement is still true if we replace $\mathbb{R}^{n\times n}$ with $X=\mathcal{C}^1(K,\mathbb{R}^{n\times n})$ equipped with the strong $\mathcal{C}^1$ topoplogy where $K\subset\mathbb{R}^m$ is compact. More precisely, is the subset $$Y=\left\{f\in X:\min_{x\in K}|\det(f(x))|>0\right\}$$ dense in $X$?
The previous proof no longer applies because if $g\in X$ the union of the singular values of $g(y)$ over $y\in K$ could be dense in a neighborhood of $0$.
There is another proof that uses that $A\mapsto \det(A)$ can be identified by a analytic map from $\mathbb{R}^{n^2}$ to $\mathbb{R}$ but I'm not sure if there is a similar idea to analytic for maps like $(f,x)\mapsto\det(f(x))$ which has an infinite dimensional domain.