Theorem 1   Let $C$ be an abelian category, and $f\colon A\to B$ be a morphism of $C$ Then $\ker f$ is an inverse limit.

Proof. Recalling the definition of a kernel in an abelian category, we can see that $i\colon X\to A$ is a kernel of $f$ if and only if the following diagrams commute: $$ \xymatrix{ & X \ar[dl]_i \ar[dr]^{f\circ i}&\\ A \ar[rr]^f & & B } $$ $$ \xymatrix{ & X \ar[dl]_i \ar[dr]^{0\circ i}&\\ A \ar[rr]^0 & & B } $$ and $f\circ i = 0\circ i$

Let $I$ be the category $$ \xymatrix{ \diamondsuit \ar@<0.5ex>[r]^\dagger \ar@<-0.5ex>[r]_\ddagger & \heartsuit } $$ and define a functor $G\colon I\to C$ by $G(\diamondsuit)=A$ $G(\heartsuit)=B$ $G(\dagger) = f$ and $G(\ddagger) = 0$

Suppose that $\liminv G$ exists and is equal to $Y$ Then there are maps $\pi_\diamondsuit\colon Y\to A$ and $\pi_\heartsuit\colon Y\to A$ that make the following diagrams commute: $$ \xymatrix{ & Y \ar[dl]_{\pi_\diamondsuit} \ar[dr]^{\pi_\heartsuit}&\\ A \ar[rr]^{G(\dagger)=f} & & B } $$ $$ \xymatrix{ & Y \ar[dl]_{\pi_\diamondsuit} \ar[dr]^{\pi_\heartsuit}&\\ A \ar[rr]^{G(\ddagger)=0} & & B } $$ This is exactly the universal condition for a kernel in an abelian category.