Let $A_*=\ldots\to A_n\to A_{n-1}\to\ldots\to A_0$ be a linear system of abelian groups. The limit of this system may be defined as the kernel of the map $$ \prod A_n\xrightarrow{g-1}\prod A_n $$ where $g$ comes from the structure maps of the system. The limit functor $\lim\colon\mathbf{Ab}^{\mathbb{N^{op}}}\to\mathbf{Ab}$ is not exact since it does not send short exact sequences in $\mathbf{Ab}^{\mathbb{N^{op}}}$ (exactness is tested in each degree) to exact sequences in $\mathbf{Ab}$ and one can therefore with the usual method construct its right derived functors $R^q\lim\colon\mathbf{Ab}^{\mathbb{N^{op}}}\to\mathbf{Ab}$ and we set $\lim^1=R^1\lim$.

On the other hand, $\lim^1$ is often defined in an other way (let me denote this object temporarily as $\lim^1'$): $$\textstyle \lim^1'A_*=coker(\prod A_n\xrightarrow{g-1}\prod A_n). $$

Why is $\lim^1 A_*=\lim^1'A_*$? I would prefer an abstract argument with derived categories, if it exists.

Since countable products 'are exact' in abelian groups, the product over $\mathbb{N}$ is already the derived product and if all the structure maps from the system would be surjective, there would be an exact triangle $$ Rlim A_*\to \prod A_n\xrightarrow{g-1}\prod A_n $$ in the derived category (of chain complexes of abelian groups). Then, the shifted $Rlim$ would equal the mapping cone of $g-1$ (which is also the homotopy cofiber of $g-1$). But this mapping cone is usually not concentrated in degree zero and in particular it is not the same as $coker(\prod A_n\xrightarrow{g-1}\prod A_n)$ which defines $\lim^1'A_*$. Hence, I don't see a connection between $\lim^1A_*$ and $\lim^1'A_*$.


This is done (unsurprisingly) in Weibel's Introduction to Homological Algebra, section 3.5. Here is the argument:

  • As hunter said, given a short exact sequence of towers $0 \to A \to B \to C \to 0$, apply the snake lemma to the following diagram to get an exact sequence $0 \to \varprojlim A \to \varprojlim B \to \varprojlim C \to \varprojlim{}^{1'} A \to \varprojlim{}^{1'} B \to \varprojlim{}^{1'} C \to 0$: $$\begin{array}{ccccccccc} 0 & \to & \prod A_i & \to & \prod B_i & \to & \prod C_i & \to & 0 \\ && \downarrow \Delta & & \downarrow \Delta & & \downarrow \Delta && \\ 0 & \to & \prod A_i & \to & \prod B_i & \to & \prod C_i & \to & 0 \end{array}$$
  • To complete the argument you need to prove that $\varprojlim{}^{1'}$ vanishes on enough injectives. Weibel constructed enough injectives of the form: $$\dots \to E = E = E \to 0 \to 0 \dots \to 0$$ where the $E$ are injective. But all the maps in this tower are surjective and so its $\varprojlim{}^{1'}$ vanishes. This proves that (in Weibel's terminology) $\varprojlim{}^{1'}$ together with $0$ in higher dimensions form a universal $\delta$-functor for $\varprojlim$, ie. $\varprojlim{}^{1'} = R^1 \varprojlim$ (and the higher ones vanish).

For more details, read Weibel's book.

  • $\begingroup$ Weibel says, "In order to show that the lim1 forms a universal $\delta$-functor, we only need to see that lim1 vanishes on enough injectives." I don't see why this is the case. Could you explain? $\endgroup$ – SorcererofDM Mar 25 '15 at 7:41

Can't notate because I'm on a tablet. Given a short exact of directed systems, the snake lemma tells us we get (functorially), an exact sequence

0 to lim A to lim B to lim C to lim 1'A to lim 1'B to lim 1'C to 0

That's enough to guarantee that lim1 and lim1' agree.

  • $\begingroup$ That's not quite enough. For example, taking the direct sum of $\operatorname{lim}^1$ with any exact functor would give a functor with the same property. $\endgroup$ – Jeremy Rickard Feb 16 '14 at 14:50
  • $\begingroup$ ah, you also need vanishing on injectives, which is true since injectives are divisible ? $\endgroup$ – hunter Feb 16 '14 at 15:17

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