I have a question about the words "acyclic" and "exact." Why does Brown use the term "acyclic" instead of "exact" in his book Cohomology of Groups? It seems to me that these two terms exactly coincide. Are there examples(or topics in math) in which being acyclic means being sth1 and being exact means being sth2, and when restricted to the homology theory sth1 and sth2 coincide? Thank you.


Acyclic and exact are not the same. As Akhil says in his answer, the long exact for group cohomology is indeed exact, but a projective resolution of a module is acyclic because it is not exact in degree zero.

Originally, one used to say that a projective resolution $P_\bullet$ of a module $M$ was "acyclic over $M$", and that means that there is a map $\varepsilon:P_0\to M$, called an augmentation, such that if one extends the complex $P_\bullet$ so as to put $M$ in degree $-1$ with $\varepsilon$ as the last differential, then the resulting complex is exact.

Similarly, an acyclic space is not one whose singular complex is exact (there are very few such spaces!) but one whose singular complex is acyclic over $\mathbb Z$.

  • $\begingroup$ What do you mean by "not exact in degree zero"? A resolution is exact by its definition. Also, what is the example of a free resolution $P$ of $M$ which is not exact? I have only Brown's book, and it seems your definitions are different. $\endgroup$ – niyazi Mar 15 '11 at 5:12
  • 1
    $\begingroup$ A resolution of $M$ is a complex whose homology is zero except in degree zero, where it is $M$. $\endgroup$ – Mariano Suárez-Álvarez Mar 15 '11 at 5:12
  • $\begingroup$ Ok, now, I see your point. You look at resolutions as separate objects. Then you attach $M$ by augmentation map. So, by combining the definition you gave above with my question, there should exists some projective resolutions which is exact but not acyclic. To get an example, we possibly need modules over non PID rings. Is the example tuff? $\endgroup$ – niyazi Mar 15 '11 at 5:51
  • $\begingroup$ Just a comment to notify you: Sándor Kovács posted an answer apparently triggered by the comments in this answer on MO. $\endgroup$ – t.b. Sep 7 '12 at 21:09

Mariano, you seem to be confusing a few things here. A complex is acyclic if and only if it is exact. (see for instance Exercise 1.1.5 in Weibel's Homological Algebra book, or probably anyplace where this is defined).

An object is acyclic for a functor if the derived functors of said functor vanish on the object. For instance a flasque sheaf for the global section functor.

A resolution is acyclic if the objects of the resolution are acyclic objects.

A projective resolution is acyclic for example for the $\mathrm{Hom}(\__, N)$ functor (for some $N$) because a projective module is and not because it is exact except at degree zero. That condition is encoded in the word resolution.

So, without knowing the way Brown uses this (niyazi, you'd have to be a little more specific about that) the answer to the question is something like this: as long as you are talking about a complex being acyclic, it means the same as exact, but acyclic also applies to an object, whereas we do not say that an object is exact.

  • $\begingroup$ See, for example, the first page of chapter V in Cartan-Eilenberg. $\endgroup$ – Mariano Suárez-Álvarez Sep 7 '12 at 21:31
  • $\begingroup$ Indeed. You are exonerated! Still, this is not the current terminology. I can offer two points: 1) Cartan and Eilenberg were the pioneers. The followers often figure out that the original terminology is not the best and history takes its course. 2) They (C&E) actually talk about left complexes, which you cannot see in newer texts and they never say exact, which was probably later introduced. $\endgroup$ – Sándor Kovács Sep 7 '12 at 21:47
  • $\begingroup$ Heh. But myguess is that Benson is using the old fashioned variants in the text that motivated this question —I don't have my copy here to check. $\endgroup$ – Mariano Suárez-Álvarez Sep 7 '12 at 21:48
  • $\begingroup$ (You mean Brown, right?). No. Here is what he says on page 5: "acyclic, i.e., $H(C)=0$". $\endgroup$ – Sándor Kovács Sep 7 '12 at 21:52

Indeed, "acyclicity" of a complex means mostly the same thing as exactness (except in degree zero, as Mariano notes). I think they are just used in different contexts: people often call sequences (such as "short exact sequences" $0 \to A \to B \to C \to 0$ or also "long exact sequences" that are derived from complexes) exact, while a complex by itself is often called acyclic if it has the same property.

For instance, one would say that the long exact sequence for group cohomology is exact (probably not acyclic), but that the resolution used to compute it was acyclic (though here "exact" is probably used more often).

  • $\begingroup$ Sorry for that! I wanted to add something at the end of my answer but I clicked the edit link for yours and added it without noticing. I rolled back my edit, and I guess/hope the end result is what you last edited. $\endgroup$ – Mariano Suárez-Álvarez Mar 15 '11 at 4:43
  • $\begingroup$ (And the little mishap ended up with me awarded a badge...!) $\endgroup$ – Mariano Suárez-Álvarez Mar 15 '11 at 4:46
  • $\begingroup$ Akhil, thank you for your comment. I gave +1. $\endgroup$ – niyazi Mar 15 '11 at 5:55

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