Is the image of the map $A \to \bigwedge^k A$ closed over $\mathbb{R}$? Let $V$ a real vector space of dimension $d$. Let $1<k < d-1$. Consider the map induced by the exterior algebra functor: 
$$ \psi:\text{End}(V) \to \text{End}(\bigwedge^kV)  \, \, \, \, , \, \, \,\psi(A)=\bigwedge^k A$$

Is the image of $\psi$ closed in the standard topology on the $\text{Hom}$-space?

 A: The answer is negative: In general $\psi({\rm End}(V))$ is not closed. Here is a proof when $d\ge4$ is even and $k=d-1$. Notice that $\Lambda^{d-1}V$ can be identified with $V$, so that $\psi(A)$ is just the cofactor matrix $\widehat{A}$.
${\rm End(V)}$ is the disjoint union of ${\rm GL}(V)$ and $\Delta$, the set defined by $\det A=0$. Because $d$ is even, $\psi$ is a homeomorphism from ${\rm GL}(V)$ onto itself, with inverse given by the formula
$$A=(\det\psi(A))^\frac1{d-1}\psi(A)^{-T}.$$
If instead $A$ is singular, then either $\psi(A)=0_V$ if $A$ has rank $\le d-2$, or $\psi(A)$ has rank $1$ if the rank of $A$ is $d-1$. In conclusion, $\psi({\rm End}(V))$ contains ${\rm GL}(V)$, which is dense in ${\rm End}(V)$, but does not contain any element of rank between $2$ and $d-1$. In particular, it is not closed.
The same analysis works also when $d\ge3$ is odd (and still $k=d-1$). One obtains that $\psi({\rm End}(V))$ contains ${\rm GL}_+(V)$, but does not contain any element of rank between $2$ and $d-1$. Hence it is not closed.
