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?
Edit:
The answer is positive if $k$ is odd:
After projecting, we obtain the following commutative diagram: \begin{matrix}\text{End}(V) &\stackrel{\psi}{\longrightarrow}& \text{End}(\bigwedge^{k}V)\\ \downarrow \pi &&\downarrow \pi\\ \mathcal{P}\big(\text{End}(V) \big)&\stackrel{\tilde \psi}{\longrightarrow}& \mathcal{P}\big(\text{End}(\bigwedge^{k}V)\big) \end{matrix}
Since $\mathcal{P}\big(\text{End}(V) \big)$ is compact, $$\pi \circ \psi \big(\text{End}(V) \big) =\tilde \psi \circ \pi \big(\text{End}(V) \big) =\tilde \psi \bigg( \mathcal{P}\big(\text{End}(V) \big)\bigg) $$ is also compact, hence closed in $\mathcal{P}\big(\text{End}(\bigwedge^{k}V)\big)$.
We now claim $\pi^{-1}\big( \pi \circ \psi \big(\text{End}(V) \big) \big)=\psi \big(\text{End}(V) \big)$ which proves $\text{Imgae}(\psi)$ is a closed set.
Clearly $\pi^{-1}\big( \pi \circ \psi \big(\text{End}(V) \big) \big) \supseteq \psi \big(\text{End}(V) \big)$. Now, let $x \in \pi^{-1}\big( \pi \circ \psi \big(\text{End}(V) \big) \big) $. Then $x=\lambda \bigwedge^k A$ for some non-zero $\lambda \in \mathbb{R}$. Since $k$ is odd, we have a $k$-th real root for every real number, so $x= \bigwedge^k (\sqrt[k]\lambda A) \in \psi \big(\text{End}(V) \big)$.
Actually I think this proof can be adapted to work also when $k$ is even: Instead of taking the standard projective space, we simply divide by a modified equivalence relation, where two elements are equivalent iff one is a positive multiple of the other. The quotient space is still compact, and now we don't have the problem of taking even real roots of negative numbers.
I think this question is equivalent to the following question:
Do there exist polynomial relations which characterize the $k$-minors of a square $d \times d$ matrix over $\mathbb{R}$? i.e. is $\psi(\text{Hom}(V,V))$ the zero set of some polynomials defined on $\text{Hom}(\bigwedge^kV,\bigwedge^kV)$?
(By "characterizing" I mean necessary and sufficient conditions for a given sequence of numbers to be the minors of some matrix. See an explicit statement below).
It seems that finding explicit relations is an open problem, e.g. see this paper. However, I am not asking about an explicit construction, only on mere existence.
Here is a more concrete phrasing:
Let $b_{(i_1,\dots,i_k),(j_1,\dots,j_k)} \in \mathbb{R}$, where $(i_1,\dots,i_k),(j_1,\dots,j_k)$ are increasing multi-indices of order $k$ ( $1 \le i_1 <i_2 < \dots<i_k\le d$).
Are there necessary and sufficient conditions for the existence of a $d \times d$ matrix $A$, such that $ b_{(i_1,\dots,i_k),(j_1,\dots,j_k)}$ is the $k$-minor of $A$, corresponding to rows $(i_1,\dots,i_k)$, and to columns $(j_1,\dots,j_k)$?
In "conditions" I mean here a set of continuous functions $\{ f_{\lambda} \}$ in the $\binom{d}{k}^2$ variables $b_{(i_1,\dots,i_k),(j_1,\dots,j_k)}$ such that the $b_{(i_1,\dots,i_k),(j_1,\dots,j_k)}$ are minors if and only if $ f_{\lambda} (b_{(i_1,\dots,i_k),(j_1,\dots,j_k)})=0$ for every $ f_{\lambda} $.