injectivity of natural map $Hom_G(V,W)\otimes V\to W$ (Kraft Procesi Exercise 3.2.4) Exercise 4a on page 27 of Kraft and Procesi's Primer on Invariant theory http://www.math.unibas.ch/~kraft/Papers/KP-Primer.pdf asks to show the following:

$V$ is an irreducible finite dimensional representation of a group $G$, with $End_G(V)=K$, the ground field, and $W$ is an arbitrary finite dimensional representation of $G$.
Then the map from $Hom_G(V, W)\otimes V$ to $W$ sending $\phi\otimes w$ to $\phi(w)$ is injective.

I wonder whether it is necessary here to require semisimplicity of $W$.  The proof I know uses this as follows:

Suppose that $\sum \phi_i(w_i) = 0$, w.l.o.g. $w_i$ linearly independent.  Then $\sum \phi_i (V)$ is a submodule of $W$ and hence has a direct sum decomposition $\oplus_j W_j$.  The composition $\pi_j\circ\phi_i: V\to V$ of every projection $\pi_j$ from this submodule to $W_j\cong V$ with $\phi_i$ must be multiplication with a scalar by the hypothesis that $End_G(V)=K$.  Because the $w_i$ are linearly independent, we can deduce for each $j$ that all these compositions are actually the zero maps, so the $\phi_i$ themselves must be zero maps.

Is semisimplicity of $W$ necessary or am I missing something else?
 A: $\newcommand{\Hom}{{\operatorname{Hom}}}$ $\newcommand{\Ker}{{\operatorname{Ker}}}$ Here is a different approach: Let $\gamma$ be the $K$-linear map $\Hom_G\left(V,W\right) \otimes V \to W, \  f \otimes v \mapsto f\left(v\right)$. It is straightforward to see that this map $\gamma$ is $G$-equivariant, where the $G$-module structure on $\Hom_G\left(V,W\right)$ is the trivial one (i.e., every element of $G$ acts as the identity). Thus,
(1) $\Ker\gamma$ is a $G$-submodule of $\Hom_G\left(V,W\right) \otimes V$.
But if $P$ is a finite-dimensional trivial $G$-module, then the $G$-module $P \otimes V$ is isomorphic to $V^{\dim P}$, and thus
(2) every $G$-submodule of $P \otimes V$ has the form $U \otimes V$ for some $K$-vector subspace $U$ of $P$.
(This follows from Exercise 3 (b) in §3.1, except that we are using different notations: e.g., our $\dim P$ corresponds to the $n$ in Exercise 3 (b), and we are tensoring with $P$ from the left instead of from the right.)
Applying (2) to $P = \Hom_G\left(V,W\right)$, we see that every $G$-submodule of $\Hom_G\left(V,W\right) \otimes V$ has the form $U \otimes V$ for some $K$-vector subspace $U$ of $\Hom_G\left(V,W\right)$. In particular, $\Ker\gamma$ has this form (because of (1)). Thus, there is a $K$-vector subspace $U$ of $\Hom_G\left(V,W\right)$ such that $\Ker\gamma = U \otimes V$. Consider this $U$.
Let $f \in U$. Every $v \in V$ satisfies $\underbrace{f}_{\in U} \otimes \underbrace{v}_{\in V} \in U \otimes V = \Ker\gamma$ and thus $\gamma\left(f\otimes v\right) = 0$, which rewrites as $f\left(v\right) = 0$ (since $\gamma\left(f\otimes v\right) = f\left(v\right)$). Hence, $f = 0$.
Let us now forget that we fixed $f$. We thus have shown that $f = 0$ for each $f \in U$. In other words, $U = 0$. Hence, $\Ker\gamma = \underbrace{U}_{=0} \otimes V = 0$, so that the map $\gamma$ is injective.
