Let $f: M_n(\mathbb{R}) \rightarrow M_n(\mathbb{R})$ defined as $f(A) = A^2$.
1) Show that $f$ is $\mathcal{C}^1$ and calculate its differential.
2) Show that there exists $g$ a $\mathcal{C}^1$ function defined on a neighbourhood of the identity matrix such that $\forall A \in V$, we have $g(A)^2 = A$
I managed to show that beginning. For a matrix $A \in M_n(\mathbb{R})$ we have $f(A) = A^2$ thus if we defined $A = (a_{i,j})_{1\leq i,j \leq n}$ we then have $f(A) = \sum^{n}_{k1}a_{i,k}a_{k,j}$. Thus it's a polynomial, thus it's $\mathcal{C}^1$ (and even $\mathcal{C}^{\infty}$).
Now, we have $f(A+H) = A^2 + AH + HA + H^2$, thus I deduce that $df(A)H = AH + HA$.
Now, for the second question: let $H \in M_n(\mathbb{R})$ then $df(I)= 2I$ thus the differential is an isomorphism, we can use the inverse function theorem. We thus have two neighbourhoods $V \subset M_n(\mathbb{R})$ and $W\subset M_n(\mathbb{R})$ such that $f: V \rightarrow W$ is a diffeomorphism of class $\mathcal{C}^1$.
Now, we can put $g = f^{-1}$ that is $\mathcal{C}^1$, and $g(A)^2 = (f^{-1})^2(A)$, and here I struggle to show that $(f^{-1})^2(A) = A$. Can someone help me out?
Note that $f(A+H)-f(A) = AH+HA+H^2$ from which we get that $Df(A)(H) = AH+HA$.
It is straightforward to see that $A \mapsto Df(A)$ is linear, hence smooth.
Note that $Df(I)(H) = 2H$, with inverse $(Df(I))^{-1} (J) = {1 \over 2} J$, hence the inverse function theorem shows that there is some $g$ defined in a neighbourood $V$ of $f(I) = I$ such that $g(f(A)) = g(A^2) = A$ for $A \in V$.