I am trying to use Algebra; Linear and otherwise to find the co-differential ,i.e., the differential
operator d in a Cohomology Theory starting with homology. For now, I just wanted to start with a
specific example I am familiar with, that of Simplicial Homology, where the differential is given
by the restriction map from an $n$-simplex S_n to all the $(n-1)-$ simplices $\{S_{n-1}\}$ incident
with $S_n$
I have some ideas, but I am far from a solution.
I know the p-th Homology group $H^{p}(M; R)$ ; M is a Topological space and $R$ is a coeficient
ring is given ( up to isomorphism) by (edit *) $$ H^P(M;R)=Hom(C_p(M;R), R)$$ , where $H_p$ is the p-th Homology group.
I am trying to make use of the dual map in linear algebra. Given (finite-dimensional, I believe)
vector spaces V,W over the same field, and a linear map L: $V \rightarrow W$ , there is a contravariant functorial
dual map between the associated duals $V^{*}, W^{*}$ given by sending $w^{*} \in W^{*}$ to $$w*L(v)$$
trying to use this dual map to define the co-differential in Simplicial Cohomology, starting
with the differential d betwen chain groups $C_n$ , given by:
restricting : an n-chain $a*c_n$ is sent to the formal sum $a*ci_{n-1}$ , where $c_{n-1}$ is
incident with $c_n$. The dual differential, aka , the co-differential would then map
from $C_{n-1}\rightarrow C_n$ through means I don't have clear. Clearly I am kind of rambling here
and would appreciate hints. Thank you.
*Subbed $C_p$ for $H_p$
Best Answer
Couple of things first
is not necessarily true, see the Universal Coefficients Theorem. Perhaps you meant to replace $H$ with $C$?
As for the intuition for $\delta$ (the co-differential as you call it), linear algebra will work, but let's think about it a little clearly.
Given $L:V \to W$, you're right that you get the dual map $L^*:W^* \to V^*$ given by $L^*(w^*) (v) = w^*(L(v))$. Essentially, you "precompose" with $L$ to give you something that an element $w^*$ can act on. The same idea works with the cochain groups.
If you have a chain complex $\rightarrow C_n \rightarrow C_{n-1} \rightarrow$, and $\partial$ is the differential (boundary operator), then note that dualizing gives you $\rightarrow C^{n-1} \to C^n \to$ where $C^\circ = \text{Hom}(C_\circ, R)$. Now note that we can define $\delta$ the same way, as $\delta(c^{n-1})(c_n)= c^{n-1}(\partial(c_n))$
In other words, we use the boundary operator to push an $n$-cycle to an $(n-1)$-cycle, and then you can use the $(n-1)$ cocycle. So you "precompose" with the boundary.