For $R$ commutative ring and $N,M$ two $R$-modules, I have the following diagram
where $(P_i,\partial_i)$ are projective modules of the projective resolution of M, $f$ is an $R$-module homomorphism and \begin{align*} &V:=\{(f(z),-\partial_1(z)):z\in P_1\}\\ &i(x):=[x,0] \qquad \qquad p([x,y]):=\partial_0(y) \end{align*} I want to prove that if $f\circ \partial_2=0$, then the sequence below is exact.
I have some trouble proving that $ \operatorname{Ker}(p)\subset \operatorname{Im}(i)$. In particular, for $[x,y]\in \operatorname{Ker}(p)$ , then $y\in \operatorname{Ker}(\partial_0)= \operatorname{Im}(\partial_1)$. Hence, there exists $z\in P_1$ such that $\partial_1(z)=y$. I want to prove that, there exists $t\in N$ such that $(x-t)=f(z)$. This is true only if $x\in \operatorname{Im}(f)$....Why $x\in \operatorname{Im}(f)?$
The map $p$ is well defined. Indeed, if $(x,y)\in V$, then $x=f(z)$ and $y=-\partial_1(z)$, so $$ \partial_0(y)=-\partial_0\partial_1(z)=0 $$ Let $[x,y]\in\ker p$, which means $\partial_0(y)=0$. The top row is exact by assumption, so $y=\partial_1(z)$.
What you want to show is that there is $x'$ such that $$ i(x')=[x',0]=[x,y] $$ that means $(x-x',y)\in V$. Thus you need $z'$ with $$ x-x'=f(z'),\quad y=-\partial_1(z') $$ An obvious candidate for $z'$ is $z'=-z$. Choose $x'=x+f(z)$ and you're done.