I have been reading about cohomology of groups. For example, consider the cohomology group $H^{1}(Z/2,Z_-)$ of Wikipedia article https://en.wikipedia.org/wiki/Group_cohomology. Here $Z/2= \{1,-1\}$ acts on $Z=\{n\}$ as, $-1.n = -n$. Wikipedia computes $H^{1}(Z/2,Z_-)$ as the group of crossed homomorphisms $\{f: Z/2\to Z\}$ satisfying $f(ab)= f(a)+ af(b)$ modulo the group of principal crossed homomorphisms $f:Z/2 \to Z; f(a)= an-n$ for some integer $n$. It is obvious from this definition that $H^{1}(Z/2,Z_-)=Z/2$.
However, the same article says that the cohomology group $H^1(G,A)$ can also be computed in the following way. One considers an exact sequence of Abelian groups $0\to A\to B \to C\to 0$ with some specified action of $G$ on $A,B$ and $C$. Then one sends each group to the group of invariants (under $G$) $A^G, B^G, C^G$ obtaining the long exact sequence $0\to A^G \to B^G\to C^G \to H^1(G,A) \to H^1(G,B)\to H^1(G,C)\to \cdots$.
I want to apply the same procedure for computing $H^{1}(Z/2,Z_-)$. In order to do that I consider the exact sequence $0\to Z \to^{d_1} Z\times Z\to^{d_2} Z\to 0$ with $d_1(n)=(n,0)$ and $d_2(n_1,n_2)=(0,n_2)$. Here $Z/2=\{1,-1\}$ acts on the left most $Z$ as $-1.n=-n$, on $Z \times Z$ as $-1.(n_1,n_2)=(-n_1,n_2)$ and trivially on the right most $Z$. Now we send each group to the group of invariants giving the exact sequence $0\to \{0\} \to \{0,n\} \to Z\to 0$. So $H^{1}(Z/2,Z_-)=0$ instead of $Z/2$.
It seems that I have misunderstood something here. It will be helpful if someone can clarify this.