This question might be too elementary but it arises naturally as a part of a more complicated computation and I struggle to find the answer.


Let $V$ be an $n$-dimensional complex vector space. Consider a map 
$$Sym^m(V)\otimes Sym^k(V)\longrightarrow Sym^{m-1}(V)\otimes Sym^{k+1}(V)$$
given by
$$x\_1 \dots x\_m \otimes y\_1 \dots y\_k \mapsto \sum x\_1 \dots x\_{i-1} x\_{i+1} \dots x\_m \otimes x\_i y\_1 \dots y\_k.$$
For $m=k=1$ this map is surjective and for $m=2,\ k=0$ it is injective. In the first case the kernel is $\Lambda^2 V$ which is also a cokernel for the second case. For $n=1$ it is always an isomorphism. 


1. For what triples of $n, m, k$ this map is injective or surjective?
2. Can the kernel / cokernel be presented functorially in terms of V?
3. Is this map a part of some natural resolution?



Maps between products of symmetric powers
[too long for a comment] Your map is the composition of two natural maps: the differential (aka polarization)
$$
Sym^m(V)\longrightarrow Sym^{m-1}(V)\otimes V
$$
(of it helps, regard $Sym(V)$ as an algebra of functions on $V^\*$), and the symmetrisation
$$
V\otimes Sym^{k-1}(V)\longrightarrow Sym^k(V)\, .
$$
The former is always injective (there are no non-constant homogeneous polynomials for $m>0$), the latter is always surjective. In order to describe its kernel for $k>2$, I guess you have to consider the map
$$
\Lambda^2(V)\otimes Sym^{k-2}(V)\longrightarrow V\otimes Sym^{k-1}(V)\, ,
$$
which is basically the odd-parity analog of your map, i.e., fist it sends $\Lambda^2(V)$ to $V\otimes V$, and then it symmetrizes $V$ with $Sym^{k-2}(V)$. [This was edited: my previous suggestion was slightly wrong]


As a matter of fact, if one of the two $V$'s appearing in your formula was a $V^\*$ instead, everything would be geometrically more transparent.


Finally, if one of the $Sym$ was a $\Lambda$ instead, then your map would be part of the Koszul complex (the de Rham differential for polynomials): I would bet that in one of these "supercommutative" geometric environments (where all the parities get mixed up) there is also a parity-altered version of the Koszul complex, which may provide you some clue about the resolution.


