5
$\begingroup$

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.

enter image description here

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

enter image description here

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!

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thank you for your answer! There is perhaps one thing I didn't quite get. What is a limit cone? And I guess that is how the $\pi_i$'s are constructed. $\endgroup$ – KittyL Nov 14 '16 at 11:56
  • 1
    $\begingroup$ @KittyL a limit cone=limit, and by a limit I mean the data composed by an object (the limit-object) and the canonical projections that give the structure of a limit to the said object. $\endgroup$ – Giorgio Mossa Nov 14 '16 at 20:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.