Let $T:V \rightarrow W$ be a linear transformation and $S \in L^k (W).$
Verify that $T^*(S^{\delta})= (T^* (S))^{\delta}, \delta \in S_k.$
Here is what I did, but unfortunately it is wrong. Please, help.
On the k-tensor powers the induced map is $T:V^{\otimes k}\to W^{\otimes k}$ which on pure tensors is $T(v_1\otimes v_2\otimes \ldots \otimes v_k)=T(u_1)\otimes T(u_2)\otimes \ldots T(u_k)$
If $\sigma$ is a permutation, then $(u_1\otimes \ldots u_k)^\sigma=u_{\sigma^{-1}(1)}\otimes \ldots \otimes u_{\otimes k}$ and thus we can immediately verify the identity
$T(u_1\otimes \ldots u_k)^\sigma=T((u_1\otimes \ldots u_k)^\sigma)$
because both sides will equal $T(u_{\sigma^{-1}(1)})\otimes \ldots T(u_{\sigma^{-1}(k)})$
Because pure tensors span the k-tensor power space, we conclude that
$T(v^\sigma)=T(v)^\sigma$ (1)
Now let's get back to the problem.
By definition, $T^*(S^\sigma)(x)=S^\sigma(T(x))$ which is $S(T(x))^\sigma$
Now by two consecutive applications of (1), $S(T(x))^\sigma=S(T(x)^\sigma)=S(T(x^\sigma))$ and that is, by definition, $(T^*S)(x^\sigma)$ - which again by definition equals $(T^*S)^\sigma(x)$