On page 192 of Hatcher's Algebraic Topology, he shows that cohomology groups for any chain complex $C_n$ fit into a short exact sequence

$$0\leftarrow \text{Ker} \,i^*_n \leftarrow H^n(C;G)\leftarrow \text{Coker} \, i^*_{n-1}\leftarrow 0$$

where $i^*_n$ is the dual of the inclusion $i_n:B_n\rightarrow Z_n$ from boundaries to cycles.

It seems to me like this dual map should always be surjective (and hence, the cokernel should always be zero). Since $B_n$ is a subgroup of $Z_n$, shouldn't any map $B_n\rightarrow G$ have an extension to $Z_n$ just by setting the map to zero for all cycles that aren't boundaries? On the same page, he makes a similar argument to show that maps out of $Z_n$ can be extended to all of $C_n$: why is the same not true for $B_n$?

  • $\begingroup$ Try this for subgroups of $\mathbb{Z}$! $\endgroup$ Feb 27, 2021 at 2:25
  • $\begingroup$ I had just done this, figured it out, and was typing up an answer as I got this notification 😅 $\endgroup$
    – Josh Ding
    Feb 27, 2021 at 2:27
  • $\begingroup$ It is a good thing to think about, so definitely not time wasted. It is one reason why working with field coefficients is easier, it is possible to do this extension when dealing with vector spaces. $\endgroup$ Feb 27, 2021 at 2:28

1 Answer 1


Oh, figured it out. Such an extension is not in general possible: I was imagining that $B_n$ and $Z_n$ were subgroups generated by individual $n$-simplices, but that's definitely not the case in general. A counterexample to the general idea is actually on the next page of Hatcher - the sequence $$0\rightarrow\mathbb{Z}\xrightarrow{n}\mathbb{Z}\rightarrow\mathbb{Z}_n\rightarrow 0$$

dualizes to

$$0\leftarrow \mathbb{Z}\xleftarrow{n}\mathbb{Z}\leftarrow 0\leftarrow 0$$

and viewing the first map in the original sequence as the inclusion of $n\mathbb{Z}$ into $\mathbb{Z}$, it's clear that the only cases where you can extend maps $n\mathbb{Z}\rightarrow\mathbb{Z}$ to all of $\mathbb{Z}$ are when those maps have degree divisible by $n$ (which is reflected in the fact that $n\mathbb{Z}$ is the image of the dualized map).


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.