# Is the inversion function $\iota:GL_{n}($R$)\rightarrow GL_{n}($R$)$ defined by $\iota(A):=A^{-1}$ continuous?

I am trying to prove the following statement:

Let $GL_{n}($R$)$ denote the set of invertible $n×n$ matrices over R.

Is the inversion function $\iota:GL_{n}($R$)\rightarrow GL_{n}($R$)$ defined by $\iota(A):=A^{-1}$ continuous?

In the first place, I was trying to use an $\epsilon$-$\delta$ argument, but I could not come up with "nice norms" to work with, or a direct relationship between the entries of one matrix and the one of its inverse.

So I decided to change strategy: I previously showed that matrix multiplication function $m:GL_{n}($R$)×GL_{n}($R$)\rightarrow GL_{n}($R$)$ is continuous and I was wondering if I can somehow use this and the fact that $\iota(A)A=I_n$ in order to finish the exercise.

I also thought that I could prove the statement using the theorem stating that if the preimage under $\iota$ of every open set is open, then $\iota$ is continuous. It seems intuitive that since $\iota$ is a bijection between the same spaces as domain and codomain, then every preimage of a open set $S\subset GL_{n}($R$)$ has the "same number of elements" of $S$ and maybe imply that $\iota^{-1}(S)$ is open as well, but I don't know how to formalize it.

Are the ideas in my strategies correct? Any hint is highly appreciated.

Each component of $A^{-1}$ is a rational function in terms of the entries of $A$, where the denominator of this rational funcion is $\det A$. Since $\det A\neq 0$, each component of the map $A\mapsto A^{-1}$ is a rational function with nonzero denominator, and is therefore continuous. Therefore, the whole map is continuous.
• Thank you! If I would like to try to go for an epsilon delta proof, how could I write more explicitly the "rational function in terms of the entries of $A$"? Commented May 6, 2017 at 20:42
• An epsilon delta proof isn't feasible, here. But what I said is sufficient, since rational functions are continuous wherever their denominators are nonzero. If you need more detail as to why the entries of $A^{-1}$ are rational functions, recall that $A^{-1} = \frac{1}{\det A}\mathrm{adj}(A)$. Commented May 6, 2017 at 20:44