The goal is to show that $\Gamma $ the set of points of continuity of $ M \mapsto \pi_M $ is $ \{ M \in M_n(k) , \chi_M = \pi_M \} $ . Where $ \pi_M $ is the minimal polynomial of $M$ and $ \chi_M $ is the characteristic polynomial of $M$.
With the field being $ k = \mathbb{R} , \mathbb{C} $ so that we can talk about topology.
$ \bullet $ First we know that $ M \mapsto \chi_M $ is continuous because Newton's identities expresses the coefficients of $ \chi_M $ as polynomial functions of $tr(A^k) $ for $ k \leq n $ and that's a finite dimensional multilinear form.
$ \bullet $ Then we see that $ M \mapsto \pi_M $ is not continuous because there exists a sequence $ (A_p)_p \subset M_n(k) $ which converges to A but $ \pi_{A_p} $ does not converge to $ \pi_A $.
$ \bullet $ So one of the two inclusions is direct. We now need to prove that :
$$ \Gamma \subset \{ M \in M_n(k) , \chi_M = \pi_M \} $$
The indication is to show that this last set in an open set of $ M_n(k) $....
Thanks in advance for any help
We work over $\mathbb{C}$. Let $A\in M_n$.
If $\chi_A\not= \pi_A$, then $A$ has a multiple eigenvalue and $\pi_A$ is a proper divisor of $\chi_A$. There is a sequence $(A_i)_i$ of matrices with distinct eigenvalues that converges to $A$. Then $\chi_{A_i}=\pi_{A_i}$ tends to $\chi_A(\not= \pi_A)$ and our function is not continuous in $A$.
If $\chi_A=\pi_A$, then $A$ admits a "cyclic vector" $u\in\mathbb{C}^n$ s.t. $u,Au,\cdots,A^{n-1}u$ is a basis of $\mathbb{C}^n$, that is, $\det(u,Au,\cdots,A^{n-1}u)\not= 0$. Let $(A_i)_i$ be a sequence that converges to $A$. Then, for $i$ great enough, by continuity of $\det$, $\det(u,A_iu,\cdots,A_i^{n-1}u)\not= 0$; thus $\chi_{A_i}=\pi_{A_i}$ (because $A_i$ admits $u$ as a "cyclic vector"; note that it's an equivalence) tends to $\chi_A(=\pi_A)$ and our function is continuous in $A$.