Let $f : A \to B$ be a homomorphism of abelian groups and $k = \text{Coker}(f, 0)$ where $0 : A \to B$ is the zero homomorphism. Then I'm trying to show that $\text{Coker}(f, 0) = (K, k)$ where $K = B/f(A)$ and $k$ is the canonical surjection.
I have that if $m : B \to M$ is another morphism such that $m\circ f = m \circ 0 = 0$ then $\ker{m} \supset f(A)$. So what theorem from abstract algebra do I use to find a factorization $n : M \to K$, $n \circ m = k$?
$B \xrightarrow{\pi} B/\ker m \xrightarrow{\sim} m(B) \xrightarrow{\iota} M$
Since $\ker m \supset f(A)$ we can define $B/f(A) \xrightarrow{\pi'} B / \ker m$ by $\pi'(\bar{a}) = \pi(a)$ and it's well defined since if $\bar{a} = \bar{b}$ then $\overline{a - b} = \bar{0}$ so that $\pi'(\overline{a-b}) = 0$.
We have the composition $\phi = \iota \circ \varphi \circ \pi' : K \to M$ as an existing arrow and we want to show that it is the unique arrow such that $\phi \circ k = m : B \to M$. Suppose that $\theta : K \to M$ is another arrow such that $\theta \circ k = m = \phi \circ k$. But this automatically implies that $\theta(\bar{b}) = \phi(\bar{b}) \ \forall \bar{b} \in K$! I guess that is obvious...