Category Theory -- limits commute with kernels

Question

This is Exercise 1.6.I from Vakil's notes on Algebraic Geometry.

Suppose $\mathscr{C}$ is an abelian category and $a: \mathscr{I}\rightarrow \mathscr{C}, b: \mathscr{I}\rightarrow \mathscr{C}$ are two diagrams in $\mathscr{C}$ indexed by $\mathscr{I}$. Let $A_i=a(i)$ and $B_i=b(i)$ be two objects in those diagrams. Let $h_{i}: A_{i}\rightarrow B_{i}$ be maps commuting with the maps in the diagram, i.e., $h_{i}$ is a natural transformation of functors $a\rightarrow b$. Then the $\ker h_i$ form another diagram in $\mathscr{C}$ indexed by $\mathscr{F}$. Describe a canonical isomorphism $\varprojlim \ker h_i\cong \ker (\varprojlim A_i\rightarrow \varprojlim B_i)$, assuming the limits exist.

I showed the first part, which is illustrated using the following diagram.

I am trying to describe the isomorphism using the following diagram:

The dashed arrows are all due to the universal property. It remains to prove there exists a unique map from $\lim(\ker h_i)$ to $\ker h$ that makes the diagram commute. Then there exists a unique isomorphism between them.

My question:

To see that, it suffices to show that the map $\lim(\ker h_i)\rightarrow \lim A_i\rightarrow \lim B_i$ is zero. I am not sure how to. Is the following argument correct?

And it is zero since it is the unique map such that the square $\lim(\ker h_i)\rightarrow \ker h_i\rightarrow B_i$ and $\lim(\ker h_i)\rightarrow \lim A_i\rightarrow \lim B_i\rightarrow B_i$ commutes. The commutativity follows from Exercise 1.3.Q. Since one side is obviously zero, the unique map makes the other side zero must be zero.

Thank you for your help!

Answer 1

I would say that you are right. Here are the details.

Let call $\psi \colon \lim(\ker h_i) \to \lim A_i$ the mapping for the limit of the $\ker h_i$ given by universal property.

You want to prove that $\phi \circ \psi=0$ the way to do that is using the following property of limits:

Assume that $(\pi_i \colon L \to B_i)_i$ is a limit cone than two morphisms $f,g \colon A \to L$ are equal if and only if for each $i$ we have $\pi_i \circ f=\pi_i \circ g$.

So to prove that $\phi \circ \psi$ is zero you just need to prove that each $\pi_i \circ \phi \circ \psi$ is equal to the zero morphism $0 \colon \lim(\ker h_i) \to B_i$.

As you observed the diagrams (one for each index $i$) $$\require{AMScd} \begin{CD} \lim(\ker h_i) @>{\pi_i}>> \ker h_i \\ @V{\psi}VV @VV{i_i}V \\ \lim A_i @>{\pi_i}>> A_i \\ @V{\phi}VV @VV{h_i}V \\ \lim B_i @>>{\pi_i}> B_i \\ \end{CD} $$ commutes, that is because of the definition of the morphisms $\psi$ and $\phi$ by universal property.

Since the right vertical composite, in the diagram above is the zero morphism, and since the external rectangle commutes, we gain that $$\pi_i \circ \phi \circ \psi = h_i \circ i_i \circ \pi_i = 0 \circ \pi_i=0\ .$$

This with the fact that $\pi_i \circ 0 = 0 \colon \lim(\ker h_i) \to B_i$, for each $i$,implies that $\phi \circ \psi=0$, as you wished.

Feel free to ask for any additional detail.