Q:

if $K\cap \text{Ker } f=0$ ($K+\text{Ker }f=M$ resp.) then $f_{\left|K\right.}:K\longrightarrow N$ is monomorphism (epimorphism resp.)

Let $f$ be $R$-epimorphism from $R$-module $M$ to $R$-module $N$ and $K$ be submodule of $M$.
Prove that if $K\cap \text{Ker }f=0$ ($K+\text{Ker }f=M$ resp.) then $f_{\left|K\right.}:K\longrightarrow N$ is monomorphism (epimorphism resp.)

A:

$f_{\mid K}$ is a monomorphism if and only if $\ker f_{\mid K} = 0$. But $\ker f_{\mid K} = \ker f$ and $K \cap \ker f = 0$. So this gives you the first claim.
For the second claim, $f_{\mid K}$ is an epimorphism, iff. $f_{\mid K}(K) = N$. But $f$ is an epimorphism $M \longrightarrow N$, so there exists a preimage $m$ for every $n \in N$ with $f(m) = n$. If $n \neq 0$ this preimage $m$ is already in $K$, since $K + \ker f = M$ and therefore $f_{\mid K}$ is an epimorphism.
Hope that helped you.

