Let $V$ be a supervector space over $\mathbb C$ and let $T^n(V):=V \otimes V \otimes \cdots \otimes V$ and let $S^n(V)$ be the super vectorspace of symmetric tensors. Then we have a cannonical surjection from $$T^n(V) \rightarrow S^n(V)$$ by the super-symmetrization map given by $$v_1 \otimes v_2 \otimes \cdots \otimes v_n \mapsto \frac{1}{n!}\sum_{\sigma \in S_n} (-1)^{sgn(\sigma)}v_{\sigma(1)} \otimes \cdots \otimes v_{\sigma(n)}$$ where $sgn(\sigma)$ is the sign obtained when $\sigma$ acts on $v_1 \otimes v_2 \otimes \cdots \otimes v_n$.
Now $\mathfrak{gl}(V)$ acts on $T^n(V)$ by a signed derivation:
$x.(v_1 \otimes v_2 \otimes \cdots \otimes v_k)= x.v_1 \otimes v_2 \otimes \cdots \otimes v_k + (-1)^{|x||v_1|}v_1 \otimes x.v_2 \otimes \cdots \otimes v_k+ \cdots +(-1)^{|x|(|v_1|+|v_2|+\cdots +|v_{k-1}|)}v_1 \otimes v_2 \otimes \cdots \otimes x.v_k,$
where $x \in \mathfrak g$ and $v_j \in V_j$ are $\mathbb Z_2$-homogeneous elements of degrees $|x|$ and $|v_j|$ respectively.
It is known that the Lie superalgebra $\mathfrak{gl}(V)$ is not reductive. Is it still true that the induced map on the invariants $T^n(V)^{\mathfrak{gl}(V)} \rightarrow S^n(V)^{\mathfrak{gl}(V)}$ is surjective ?
