I'm trying to learn more about cohomology in general. I have a decent understanding of homology using simplicial complexes so building chain groups as free abelian groups generated by simplices and boundary operators being alternating sums of a simplices faces all makes intuitive sense to me. But when I read more about cohomology I'm struggling to grasp what exactly the coboundary operator is. I understand the the cochain groups $C^n$ are groups of homomorphisms from the chain groups $C_n$, that is $C^n = Hom(C_n,R)$ for some ring $R$. But how exactly do we compute the coboundary operator?

I found this source which has a great explanation except I can't seem to understand why the coboundary map in 5.8.4 sends elements from $C^{n-1}$ to $C^n$. If we remove a an index, $r$, of a simplex $I \in N^{k-1}$ denoted as $I_r$ doesn't it become a simplex in $N^{k-2}$ not $N^k$?

I feel like there is some implicit concept I don't fully understand so any help would be appreciated :)

  • $\begingroup$ $\delta c(-)$ takes elements in $N^{k}$ and outputs an element of $\mathbb{R}$, I had trouble with understanding this aswell so don't worry if it doesn't click right away. Also if you know about simplicial sets you should look into the Cech nerve of an open covering. Then the Cech cohomology of that open covering is just the cohomology of the Cech nerve with coefficients in $\mathbb R$. $\endgroup$ Commented Nov 13, 2019 at 10:44

1 Answer 1


For what it's worth, I like to think of this issue in a culinary fashion.

An $n$-dimensional cocycle $c$ eats $n$ dimensional simplices $\sigma$ and spits out numbers $c(\sigma)$. More generally, $c$ can also eat an $n$ dimensional chain $\sum_i a_i \sigma_i$: it does this by separately eating each $\sigma_i$, spitting out the numbers $c(\sigma_i)$, and then forming the linear combination $\sum_i a_i c(\sigma_i)$, so $$c \left(\sum_i a_i \sigma_i \right) = \sum_i a_i c(\sigma_i) $$

Now, given an $n-1$ dimensional cochain $c \in C^{n-1}$, we want to define the $n$-dimensional cochain $\delta c \in C^n$.

What does $\delta c$ eat? It eats an $n$-dimensional simplex $\sigma$.

What does $\delta c$ spit out? It spits out a number $\delta c(\sigma)$. And the value of that number it spits out is defined to be $$\delta c(\sigma) = c(\partial \sigma) $$ And does that make any sense? Yes it does: $\sigma$ is an $n$-dimensional simplex, so $\partial\sigma$ is an $n-1$ dimensional chain, and that's exactly what the $n-1$ dimensional cochain $c$ likes to eat.

  • 2
    $\begingroup$ This is great! Thank you so much. After playing around more with this I've decided that I really like this notation: $\delta_{n}(c) = c \circ \partial_{n+1}$ because the coboundary operator eats a function and spits out a function and that function goes from eating $n$-chains to eating $n+1$-chains. $\endgroup$
    – markagrios
    Commented Nov 14, 2019 at 14:51

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