I have the following situation:
I have this commutative diagram of $R$-modules and homomorphisms such that each row is exact. First I want to show that if $f$ and $h$ are monomorphisms then $g$ is a monomorphism.
To do so I wanted to show that $\ker(g)=\{0\}$ since we just know that $g$ is a homomorphism. Therefore I took $x\in \ker(g)$, so $g(x)=0_N$. But since the rows are exact I know that $N'\rightarrow N$ needs to be injective. So $\bar u(0)=0_N=g(x)$ (I denote now $u$ tilde as $\bar u$). But then I don't see where to continue, could maybe someone help me?
Let $x\in\ker(g)$, so $g(x)=0$. Then, $h\circ v(x)=\tilde v\circ g(x)=0$, so $v(x)\in\ker(h)$. But since $h$ is injective, $v(x)=0$, so $x\in\ker(v)=\mathrm{im}(u)$. Let $x=u(y)$ for some $y\in M'$. Then, $\tilde u\circ f(y)=g\circ u(y)=g(x)=0$, so since both $\tilde u$ and $f$ are injective, we have $y=0$.
Thus, $x=u(y)=0$.