How can I show that the map $f: GL_n(\mathbb R)\to GL_n(\mathbb R)$ defined by $f(A)=A^{-1}$ is continuous?
The space $GL_n(\mathbb R)$ is given the operator norm and so I want to show for all $\epsilon$ there exists $\delta$ such that $\|A-B\|<\delta \implies \|A^{-1}-B^{-1}\|<\epsilon$.
On
In fact is true in any unital Banach algebra. See Lemma 5.1 of https://www.math.ksu.edu/~nagy/real-an/2-05-b-alg.pdf or Banach Algebras: Continuity of Inversion?.
On
C.Falcon gave the simplest and most elegant answer with the algebraic formula involving the determinant. That formula actually implies that $A^{-1}$ is an analytic function of $A$. We can prove this fact with a direct, determinant-free proof that extends to the infinite dimensional case (see also the answer of Martín-Blas).
The key step is the geometric series formula $$ f(I-Y)=(I-Y)^{-1}=\sum_{n=0}^\infty Y^n, $$ which holds for all matrices $Y$ such that $\|Y\|<1$, where $\|\cdot\|$ is a matrix norm that we fix once and for all. This shows that $f$ is analytic in a neighborhood of $I$. If $M$ is an invertible matrix, then $$ f(M-X)=M^{-1}(I-M^{-1}X)^{-1}=M^{-1}\sum_{n=0}^\infty (M^{-1}X)^n, $$ provided that $$\tag{*} \|M^{-1}X\|<1.$$ Since $\|\cdot\|$ is a matrix norm it satisfies the inequality $\|AB\|\le \|A\|\|B\|$, and in particular, the property (*) is satisfied if $$ \|X\|<\frac{1}{\|M^{-1}\|}.$$ We conclude that $f$ is analytic in a neighborhood of $M$, and so it is analytic on the whole set of invertible matrices.
In particular, $f$ is continuous.
On
The crucial inequality is this one:
$$\|A^{-1}-B^{-1}\|=\|-A^{-1}(A-B)B^{-1}\|\le\|A^{-1}\|\|A-B\|\|B^{-1}\|\le \|A^{-1}\|\|A-B\|(\|A^{-1}\|+\|A^{-1}-B^{-1}\|)$$
The rest is just finding how small to make $\|A-B\|$ to make $\|A^{-1}-B^{-1}\|\lt\varepsilon$. I've done this calculation, and it turns out that it is enough to take:
$$\delta=\min\left(\frac{1}{\|A^{-1}\|},\frac{\varepsilon}{\|A^{-1}\|^2+\|A^{-1}\|\varepsilon}\right)$$
Namely, if $\|A-B\|\lt\delta$, then $\|A-B\|\lt\frac{1}{\|A^{-1}\|}$, i.e. $1-\|A^{-1}\|\|A-B\|\gt 0$, and also: $\|A-B\|\lt\frac{\varepsilon}{\|A^{-1}\|^2+\|A^{-1}\|\varepsilon}$, i.e. $\|A^{-1}\|^2\|A-B\|+\|A-B\|\|A^{-1}\|\varepsilon\lt\varepsilon$, i.e. $\|A^{-1}\|^2\|A-B\|\lt (1-\|A^{-1}\|\|A-B\|)\varepsilon$, i.e.
$$\frac{\|A^{-1}\|^2\|A-B\|}{1-\|A^{-1}\|\|A-B\|}\lt\varepsilon$$
On the other hand, if you define $h=\|A^{-1}-B^{-1}\|$, then the first inequality above can be rewritten as:
$$h\le\|A^{-1}\|\|A-B\|(\|A^{-1}\|+h)$$
i.e. $h(1-\|A^{-1}\|\|A-B\|)\le\|A^{-1}\|^2\|A-B\|$, i.e. $h\le\frac{\|A^{-1}\|^2\|A-B\|}{1-\|A^{-1}\|\|A-B\|}$.
Thus $\|A^{-1}-B^{-1}\|=h\lt\varepsilon$, as needed.
On
Use the fact that $A^{-1} = \frac{1}{\textrm{det(A)}}\textrm{adj}(A)$ and the following general principles:
1 . The topology on $\textrm{GL}_n(\mathbb{R})$, with the operator norm, is the subspace topology coming from $\mathbb{R}^{n^2}$, the $n^2$-fold product of copies of $\mathbb{R}$. In general, norm topology on finite dimensional vector spaces coincides with product topology.
2 . If $Z$ is a subspace of a topological space $Y$, then a function $f: X \rightarrow Z$ is continuous if and only if $f$ is continuous as a map from $X$ to $Y$. This follows directly from the definition of the subspace topology. Similarly, if $S$ is a subspace of $X$, and $f: X \rightarrow Y$ is a continuous function, then the restriction of $f$ to $S$ is a continuous function from $S$ to $Y$.
3 . If $Y$ is a topological space, and $n$ is an integer, let $Y^n$ be the $n$-fold product of copies of $Y$. Let $\pi_i: Y^n \rightarrow Y$ be the function $(y_1, ... , y_n) \mapsto y_i$. A function $f: X \rightarrow Y^n$ is continuous if and only if each $\pi_i \circ f: X \rightarrow Y$ is continuous. This follows directly from the definition of the product topology.
Using these principles, you can reduce the problem to verifying that certain functions going into $\mathbb{R}$ are continuous. For example, the determinant function $\textrm{det}$ is continuous, and $\frac{1}{\textrm{det(g)}}$ is a continuous function $\textrm{GL}_n(\mathbb R) \rightarrow \mathbb R$, because it is the composition of the determinant function and the function $\frac{1}{x}$.
On
One can use the Cayley-Hamilton Theorem, which states that if $$p(x)=\det(xI-M)= x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$$ is the characteristic polynomial of $M$, then
$$M^n+a_{n-1}M^{n-1}+\cdots+a_1M+a_0I=0$$.
If $M\in GL_n(\mathbf R)$, then note that $a_0$ above cannot be $0$, for otherwise the determinant of the matrix $M$ would be $0$. Also note that each $a_i$ is a polynomial in the entries of $M$. So we can think of each $a_i$ as a map $GL_n(\mathbf R)\to \mathbf R$, which takes a matrix $M$ to $a_{i, M}$, and we know that this map is continuous.
Now note that for any matrix $M\in GL_n(\mathbf R)$, by the Cayley-Hamilton Theorem stated above,
$$\frac{-(M^{n-1}+a_{n-1, M}M^{n-2}+ \cdots+ a_{2, M}M+ a_{1, M}I)}{a_{0, M}}$$
is the inverse of $M$.
So if we define the map $f:GL_n(\mathbf R)\to GL_n(\mathbf R)$ as
$$f(M)= \frac{-(M^{n-1}+a_{n-1, M}M^{n-2}+ \cdots+ a_{2, M}M+ a_{1, M}I)}{a_{0, M}}$$
then by the remarks above it is clear that $f$ is continuous and it takes $M$ to $M^{-1}$.
One has a formula for the inverse, namely: $$A^{-1}=\frac{1}{\det(A)}\operatorname{adj}(A),$$ where $\operatorname{adj}(A)$ is the adjugate of $A$. Whence, $A^{-1}$ is a rational fraction in the coefficients of $A$.
Another approach could be the following, though it is not exactly rigorous as it assumes that if $(A_k)_k$ converges, then the sequence $({A_k}^{-1})_k$ converges too.
Let $A$ be invertible and let $(A_k)_k$ be a sequence of invertible matrices converging toward $A$, then one has: $${A_k}{A_k}^{-1}=I_n,$$ therefore, by bilinearity of the matrix product and hence its continuity, one has: $$A\lim_{k\to+\infty}{A_k}^{-1}=I_n,$$ so that by unicity of the inverse, one has $\lim\limits_{k\to+\infty}{A_k}^{-1}=A^{-1}$, which is the desired result.