I am learning local cohomology from Hartshorne’s Local Cohomology book. My question is about the notion of essentially zero inverse system of abelian groups, which is defined to be an inverse system of abelian groups $(M_{m})$ indexed by the non-negative integers, such that for every $m$, there is some integer $m’\geq m$, such that $M_{m’}\rightarrow M_{m}$ is the zero map.

Hartshorne’s book has the following remark, which I have trouble verifying:

If we have an exact sequence of inverse systems

$$0\rightarrow (M’_{m})\rightarrow (M_{m})\rightarrow (M''_{m})\rightarrow 0,$$

then the middle one is essentially zero if and only if the two outside ones are essentially zero.

I could only show that if the middle inverse system is essentially zero, then so are the two outside inverse systems.

The idea is to use a commutative diagram of the two short exact sequences induced by the zero map $M_{m’}\rightarrow M_{m}$.

So my question is, how to show that if the two outside inverse systems are essentially zero, then so is the middle one? I have tried to apply the same idea as above, but did not get a proof. Thank you so much for your kind help.

`^{\prime\prime}`

, if for some reason you like that better). Compare $M”_m$`M”_m`

to $M''_m$`M''_m`

. I edited accordingly. $\endgroup$1more comment