Define $S=\{f\in\mathcal{C}^1([0,1]^d,\mathbb{R}^{d\times d}):\det(f(x))\text{ is tranverse to }0\}$
Another way of defining $S$ is $\{f\in\mathcal{C}^1([0,1]^d,\mathbb{R}^{d\times d}):\text{for all }x\in[0,1]^d,\ \det(f(x))\ne 0\text{ or }\nabla\det(f(x))\ne0\}$.
I was able to show that $S$ is open and I am hoping to show that $S$ is dense.
This is my attempt for showing that $S$ is dense. Fix $g\in\mathcal{C}^1([0,1]^d,\mathbb{R}^{d\times d})$. By the transversality lemma, for almost every $\theta\in\mathbb{R}$, $\det(f(x))+\theta$ is transverse to $0$. In particular, there exists a sequence of numbers $\theta_n$ in $\mathbb{R}$ converging to $0$ such that $\det(g(x))+\theta_n$ is tranverse to $\{0\}$. Then I tried constructing a function $g_n\in\mathcal{C}^1([0,1]^d,\mathbb{R}^{d\times d})$ such that $g_n$ converges to $g$ and $\det(g_n(x))=\det(g(x))+\theta_n$ but this is not possible by this answer. Is there another proof approach or is the statement false?