I'm going through lecture notes on categories and homological algebra. There seems to be a very basic fact that's eluding me.
Let $C$ be a category and $f,g\in Hom(X_0,X_1)$, such that $K=\ker(f,g)$ exists, then one gets a map $i:K\to X_0$. This map is a monomorphism. The proof of that fact goes as follows :
Take a pair of parallel arrows $h,k: Y \rightrightarrows K$ such that $ih=ik=w$ then $fw=fih=gih=gik=gw$. Hence $w$ factors uniquely through $i$, thus $h=k$.
I get that $w$ factoring uniquely through $i$ implies that $h=k$, I don't get why that string of equality implies $w$ does indeed factor uniquely through $i$.
Keep in mind, that the kernel is a special limit, namely that of the diagram $f,g\colon X_0\rightrightarrows X_1$. According to the universal property of the limit, any morphism $w\colon Y\rightarrow X_0$, that fulfills $f\circ w=g\circ w$ (therefore is a cone of the diagram), factors uniquely over their kernel (limit of the diagram) $K=\ker(f,g)$:
In your proof, the morphism $w=ih=ik\colon Y\rightarrow X_0$ fulfills the condition $f\circ w=g\circ w$ of being a cone of the diagram $f,g\colon X_0\rightrightarrows X_1$, which is exactly what the string of equalities shows.