the determinant function is an open function? The determinant function $\det:M(n,\mathbb R)\rightarrow \mathbb R$ is an open mapping or a closed mapping?
The determinant function $\det:M(n,\mathbb C)\rightarrow \mathbb C$ is an open mapping or a closed mapping?
 A: In the complex case we can use this lemma: Let $P$ be a nonconstant polynomial in $z_1,\dots ,z_m.$ Then $P$ is an open map of $\mathbb {C}^m$ to $\mathbb {C}.$ Since the determinant is a nonconstant polynomial on $\mathbb {C}^{n^2},$ the desired result for $M_n(\mathbb {C})$ follows. 
Proof of lemma: Let $B(a,r)$ be an open ball in $\mathbb {C}^m.$ Then $P$ cannot be constant on $B(a,r)$ (this is true of any nonconstant polynomial on any euclidean space). For $w\in \mathbb {C}^m,$ let $P_w(\lambda)= P(a+\lambda w)$ for $\lambda \in \mathbb {C}.$ Then each $P_w$ is a holomorphic polynomial on $\mathbb {C}.$ We have
$$P(B(a,r)) = \cup_{\|w\|=1} \{P(a+\lambda w): |\lambda| < r\}.$$
By the open mapping theorem in one complex variable, each $P_w$ is either constant or an open map. Thus each set in the union above is either $\{P(a)\}$ or an open set containing $P(a).$ Because $P$ is nonconstant on $B(a,r),$ some of the sets in the union are open, hence the union is open. This shows $P$ is open as desired.

I thought I'd add this proof for the real case: Claim: $\det$ has no local extrema. Suppose we show this. Let $B$ be an open ball in $M_n(\mathbb {R}).$ Because $\det$ is continuous and $B$ is connected, $\det B$ is a bounded interval. This interval cannot take the form $[a,b),[a,b],(a,b],$ otherwise $\det$ has an extremum in $B.$ Thus every $\det (B)$ is an open interval, hence $\det$ is an open map.
Proof of claim: Suppose $\det M \ne 0.$ (This is the easy case.) Let $r>0.$ The map $f(x) =x M$ is continuous from $\mathbb {R}$ to $M_n(\mathbb {R}).$ Thus there is an open interval $I$ about $1$ such that $f(I)\subset B(M,r).$ For these $x$ we have $\det (xM) = x^n \det M,$ from which we can see $\det [B(M,r)]$ includes values above and below $\det M.$ Thus $\det$ does not have a local extremum at $M.$
Now suppose $\det M =0.$ Let $S$ be the linear transformation whose matrix is $M$ (relative to the standard basis). Then we can write $$\mathbb {R}^n = X\oplus \ker S = S(\mathbb {R}^n) \oplus Y,$$ where $m=\dim X = \dim S(\mathbb {R}^n)$ for some $m<n,$ and $\dim \ker S = \dim Y=n-m.$ Let $T$ be a linear transformation such that $T = 0$ on $X,T(\ker S) = Y.$ Then $S+xT$ is nonsingular for $x\in \mathbb {R},x\ne 0.$ Hence $\det (S+xT)\ne 0$ for such $x.$ To make sure we have both positive and negative values, let $u_1,\dots, u_m$ be a basis for $X,$ $u_{m+1},\dots u_n$ a basis for $\ker T.$ Let $T'$ be the linear transformation such that $T'=T$ on this basis except for $u_n,$ where we define $T'(u_n) = - T(u_n).$ Then $\det (S+xT') = -\det (S+xT),$ which you can see by looking at the matrices of these transformations with respect to the basis $u_1,\dots ,u_n.$
If we now let $M(x),M'(x)$ be the matrices of $S+xT, S+xT'$ with respect to the standard basis, we see that $\det$ takes on positive and negative values in any $B(M,r).$ That finishes the proof.
A: To complement Zach Stone's nice answer, notice also that $\det$ is not a closed map for $n\ge2$ (if $n=1$, then $\det = \operatorname{id}$). Indeed, in both cases you can consider the set of matrices that are given by
$$M_t=\pmatrix{t&0\\0&\frac{1}{t^2}}$$
in the upper left corner and correspond with the identity matrix everywhere else for $t$ running on $(0,\infty)$. It is closed, as we have
$$\|M_t\| = \cases{t&if $t\ge1$\\ \frac{1}{t^2}&if $t\le1$}$$
and thus any convergent sequence in this set must be of the type $\{M_{t_n}\}_n$ with $\{t_n\}_n$ converging to a finite, non-zero number (else the norm would diverge). However, we have that
$$\det(\{M_t|t\in(0,\infty)\}) = (0,\infty),$$
which is not closed.
A: It is an open function! First of all, We will state a standard result about open functions. Suppose $f:X\to Y$. Then $f$ is open if and only if for every point $x$, and every neighborhood $U$ of $x$, there is a neighborhood $V$ of $f(x)$ so that $V \subset f(U)$. Furthermore, it suffices to assume $U$ is basic open.
I don't know how to prove this simultaneously for invertible and non-invertible matricies, so we'll have to do cases. Sorry it's so long... I'll keep thinking about a simpler solution.
Let's start with the invertible case. Let $\varepsilon >0$. Assume $M$ is an invertible $n\times n$ matrix over $\mathbb{F}$ ($\mathbb{F}$ can be complex or real, it doesn't matter). Then fix a small $x$. (We will figure out how small later.) In the real setting, let $z$ be the $n^\text{th}$ root of $1+x$. in the complex setting we need the root with the smallest argument. I.E.
$$z = \sqrt[n]{|1+x|}e^{(\arg 1+x)/n}$$
Now, assume $x$ is small enough so that
$$
|1-z|<\frac{\varepsilon}{\|M\|}
$$
So we have
$$
\|M-zM\|\leq |1-z|\;\|M\|<\varepsilon
$$
Lastly,
$$
\det(zM)=z^n\det(M)=(1+x)\det M
$$
Now, because $\det M \neq 0$, we can choose $x$ to give us any value nearby $\det M$. So there is some neighborhood $V$ around $\det M$ with $V\subset f[B(M,\varepsilon)]$
We can now get to work on the non-invertible case. Let $\varepsilon >0$. Suppose $M$ is a non-invertible $n\times n$ matrix. Let $M=SJS^{-1}$ be the Jordan decomposition of $M$. Then $J$ is a Jordan form matrix, and we know that at least $1$ of it's diagonal entries is $0$. Let $\lambda_1,...,\lambda_i$ be the non-zero eigenvalues. (There many not be any, just let $i=0$ in that case.) Then there are $j:= n-i$ many $0$ diagonal entries. Let $J'$ be the matrix where the first $0$ diagonal is replaced by $x$, and the rest by $|x|$ for some small $x$. ($M$ is singular so there is at least one such entry). Observe
$$
\|M-SJ'S^{-1}\|=|x|
$$
Also,
$$
\det(SJ'S^{-1})=\det J'= x|x|^{j-1}\Pi_{k=0}^i \lambda_k 
$$
So we want $|x|<\varepsilon$. So by choosing $x$ appropriately, we can get a small neighborhood around $0$, call it $V$, where $V\subset f[B(M,\varepsilon)]$
Then then we're done! yay.
A: Here's another approach for the invertible case:
Fix an invertible matrix $A$, and let $B$ be any matrix, then
$$
\frac{d}{dt}\biggr|_{t=0} \det(A+tAB) = \det(A)\frac{d}{dt}\biggr|_{t=0} \det(I+tB) = \det(A)\mathrm{tr}(B)
$$
Since we can pick matrices of arbitrarily small norm with positive and negative traces, we get that $\det$ is an open map about $A$.
