(Hatcher Section 3.3, page 243) First, recalling the definition of a directed system of groups:

Suppose one has abelian groups $G_\alpha$ indexed by some partial ordered index set $I$ having the property that for each pair $\alpha, \beta \in I$ there exists $\gamma \in I$ with $\alpha \leq \gamma$ and $\beta \leq \gamma$. Such an $I$ is called a directed set. Suppose also that for each pair $\alpha \leq \beta$ one has a homomorphism $f_{\alpha\beta} : G_\alpha \to G_\beta$, such that $f_{\alpha\alpha} = 1$ for each $\alpha$, and if $\alpha \leq \beta \leq \gamma$ then $f_{\alpha\gamma}$ is the composition of $f_{\alpha\beta}$ and $f_{\beta\gamma}$.... this data... is called a directed system of groups.

Hatcher then gives two alternate definitions for the direct limit group $\varinjlim G_\alpha$ for a directed system of groups $\{G_\alpha\}_{\alpha \in I}$.

"The shorter definition":

... the quotient of the direct sum $\bigoplus_\alpha G_\alpha$ by the subgroup generated by all the elements of the form $a - f_{\alpha\beta}(a)$ for $a \in G_\alpha$, where we are viwing each $G_\alpha$ as a subgroup of $\bigoplus_\alpha G_\alpha$.

and "the other definition, which is often more convenient to work with":

Define an equivalence relation on the set $\bigsqcup_\alpha G_\alpha$ by $a \sim b$ if $f_{\alpha\gamma}(a) = f_{\beta\gamma}(b)$ for some $\gamma$, where $a \in G_\alpha$ and $b \in G_\beta$.... It could also be described as the equivalence relation generated by setting $a \sim f_{\alpha\beta}(a)$. Any two equivalence classes $[a]$ and $[b]$ have representatives $a'$ and $b'$ lying in the same $G_\gamma$, so define $[a] + [b] = [a' + b']$. One checks that this is well-defined and gives an abelian group structure to the set of equivalence classes. It is easy to check further that the map sending an equivalence class $[a]$ to the coset of $a$ in $\varinjlim G_\alpha$ is a homomorphism, with an inverse induced by the map $\sum_i a_i \mapsto \sum_i [a_i]$ for $a_i \in G_\alpha$. Thus we can identify $\varinjlim G_\alpha$ with the group of equivalence classes [a].

Now, Exercise 17 in that section (page 259) asks,

Show that a direct limit of exact sequences is exact.

The problem continues with

More generally, show that homology commutes with direct limits: If $\{C_\alpha, f_{\alpha\beta}\}$ is a directed system of chain complexes, with the maps $f_{\alpha\beta} : C_\alpha \to C_\beta$ chain maps, then $H_n(\varinjlim C_\alpha) = \varinjlim H_n(C_\alpha)$.

For now, I'd like to focus on the first part of the exercise.

I am having the problem I often have with this text, which is that when I read a longer passage informally giving a definition or a structure, I feel like I follow it fine, sentence by sentence, while reading it - but then, at the end of the paragraph, I don't have a grasp of what was just discussed.

I feel as though I can individually follow each of the definitions given above, but in response to "Show that the direct limit of exact sequences is exact," I draw a blank.

My question here: What is a "direct limit of exact sequences"? How do I write it down? (From there, I imagine that showing that structure, whatever it is, is exact, will follow straightforwardly.)


A direct limit of exact sequences is like a direct limit of groups, except you replace each individual group by an exact sequence. So you have a direct set $I$, and for each $\alpha\in I$ you have an exact sequence $A_\alpha\stackrel{s_{\alpha}}{\to}B_\alpha\stackrel{t_\alpha}{\to} C_\alpha$, and whenever $\alpha<\beta$ you have maps $f_{\alpha\beta}:A_\alpha\to A_\beta$, $g_{\alpha\beta}:B_\alpha\to B_\beta$, and $h_{\alpha\beta}:C_\alpha\to C_\beta$, such that $g_{\alpha\beta}s_\alpha=s_\beta f_{\alpha\beta}$ and $h_{\alpha\beta}t_\alpha=t_\beta g_{\alpha\beta}$ (i.e., the obvious diagram commutes). Furthermore, we require that $f_{\alpha\gamma}=f_{\beta\gamma}f_{\alpha\beta}$ and similarly for $g$ and $h$, so we have three separate directed systems of groups.

You can then take the direct limits $\varinjlim A_\alpha$, $\varinjlim B_\alpha$, and $\varinjlim C_\alpha$, and you can show that the maps $s_\alpha$ and $t_\alpha$ induce maps $s:\varinjlim A_\alpha\to \varinjlim B_\alpha$ and $t:\varinjlim B_\alpha\to \varinjlim C_\alpha$. The question is then asking you to show that $s$ and $t$ also form an exact sequence.

To put it another way, directed systems of groups indexed by $I$ form a category: a map between a directed system $(A_\alpha, f_{\alpha\beta})$ and a directed system $(B_\alpha,g_{\alpha\beta})$ consist of a map $s_\alpha:A_\alpha\to B_\alpha$ for each $\alpha$ such that $g_{\alpha\beta}s_\alpha=s_\beta f_{\alpha\beta}$ for all $\alpha\leq\beta$. Such a map induces a map between the direct limits (i.e. taking the direct limit is a functor from the category of directed systems to the category of groups). You can then define an exact sequence of directed systems of groups to be a sequence of maps of directed system such that for each $\alpha$, the $\alpha$-terms of the sequence form an exact sequence of groups. The question is then whether given an exact sequence of directed systems, the induced sequence on their direct limits is also exact.

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  • $\begingroup$ Thank you for the response. I think I follow the logic here (aside from the category theoretic viewpoint, as I haven't studied any category theory yet). As for the generalization to "homology commutes with direct limits", is it not essentially a one-liner, given the initial result that direct limits of exact sequences is exact? My intuition says that yes the result follows almost trivially, but I'm not sure where I'd begin with a formal proof if asked (meaning, of course, that it isn't trivial). $\endgroup$ – user169845 Mar 1 '16 at 3:17
  • $\begingroup$ Yes, it does follow essentially "trivially", though you might find it easier to just prove it directly instead of deducing it from the first question. The point is that saying "$H$ is the homology of $A\to B\to C$" is the same as saying that there are a bunch of exact sequences with certain properties (the idea is similar to my answer here). If you want more details, I suggest you ask that as a separate question. $\endgroup$ – Eric Wofsey Mar 1 '16 at 3:24

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