The inverse does not extend continuously across singular matrices

Idea. A continuous map is bounded on a neighbourhood of any point, but the inverse blows up as we approach a singular matrix.

Step 1: suppose $F$ exists. Then $F$ is continuous at $0$, so there are $\delta>0$ and $M<\infty$ with $\|F(A)\|\le M$ whenever $\|A\|<\delta$ (norm of your choice).

Step 2: test matrices. For $0<\varepsilon<\delta/\sqrt n$ the matrix $\varepsilon I$ is invertible with $\|\varepsilon I\|_F=\varepsilon\sqrt n<\delta$, and
$$\|F(\varepsilon I)\|_F=\|\varepsilon^{-1}I\|_F=\frac{\sqrt n}{\varepsilon}.$$
This tends to $+\infty$ as $\varepsilon\to0^+$, contradicting $\|F(\varepsilon I)\|\le M$.

Remark. The same argument works at any singular matrix $A_0$. Choose $x\ne0$ with $A_0x=0$. Then $(A_0+\varepsilon I)x=\varepsilon x$, so for small $\varepsilon\ne0$ (where $A_0+\varepsilon I$ is invertible) $(A_0+\varepsilon I)^{-1}x=x/\varepsilon$, hence $\|(A_0+\varepsilon I)^{-1}\|_{op}\ge1/\varepsilon\to\infty$. So the inverse cannot be extended continuously to any point of the singular set: $GL_n(\mathbb{R})$ is the maximal domain on which inversion is continuous (in the sense of extension to a larger set of limit points).

Lesson. Unboundedness near a point kills any continuous extension.