Prove or disprove that given two G-representations V,W over $\mathbb{c},$ there is an injective morphism of G-representations $\theta : W\to V$ iff there is a surjective morphism of G-representations $\theta' : V\to W$.
Below is my (incomplete) attempt at proving the forward direction.
Suppose there exists an injective morphism $\theta : W\to V$ and suppose the two G-representations are $\rho$ and $\phi.$ Then for any $g\in G,\theta \rho_g = \phi_g \theta.$ Because of injectivity, any $w\in W$ is mapped to a unique $v := \theta(w)\in V$, so let $\theta'(v) = w$ if $\theta(w)=v$ for some w and let $\theta'(v) = 1$ otherwise. Then $\theta'$ is clearly surjective and well-defined. We need to show that for any $v\in V, \theta' \phi_g(v) = \rho_g \theta'(v),$ but I don't think this holds. I'm not exactly sure how to define $\theta'$ for values of v not in the image of $\theta.$
This is wrong. Consider $G=\Bbb Z$ and let $V$ be the regular $G$-representation over $\Bbb C$. Let $W$ be the trivial $G$ representation on a one-dimensional vector space. There's a surjective homomorphism of representations: $V \to W$ (just map a free generator of $W$ as a $\Bbb C[G]$-module to any non-zero vector in $W$). But if there was any nonzero homomorphism $W \to V$, we would have $V^G\neq 0$, but in fact we have $V^G=0$.