I recently asked this question. Now, the answer there claimed that the functor $()^G:Rep_G\to Vect_{\mathbb{C}}$, where $Rep_G$ are complex representations of a group $G$, and $V^G=\{v\in V: \rho(g)v=v\forall g\in G\}$, is left exact since it is representable as a Hom functor, so left exactness holds. This is all well and good (a bit above my level but interesting nonetheless) but this question is asking about a specific counterexample.
Let $0\to A\to B\to C\to 0$ be an exact sequence, $G=\mathbb{Z}$ and let $B=\mathbb{C}^2$ with $\rho_B(n)=\begin{pmatrix} 1 & n\\ 0 & 1\end{pmatrix}$. I'm supposed to show that $B^G\to C^G\to 0$ is not exact.
So, $B^G$ is $1$ dimensional spanned by $\begin{pmatrix} 1\\ 0\end{pmatrix}$, so if we suppose $S:B\to C$ in the exact sequence, and $S^G:B^G\to C^G$ in the corresponding sequence makes the sequence $0\to A^G\to B^G\to C^G\to 0$ exact. Then either $\dim C^G=0,1$, and they both reduce to the same case: either $S^G$ is an isomorphism or $T^G:A^G\to B^G$ is an isomorphism, for $T:A\to B$ in the original sequence. Now this is where I'm stuck.
Ideas?
Edit: Many proofs of this that I've seen (such as that in Rotman's Homological Algebra book) just rely on the left exactness of the hom functor, which is fine but not helpful to understanding this specific example. Edit 2: I did prove that $0\to A^G\to B^G\to C^G$ without much difficulty by direct computation.