Cochain complex as the dual

Transformations of the spaces given by chain complexes are captured by linear operators. Meanwhile scalar quantities that depend on the location in a space given by a chain complex are linear functionals.

To illustrate, an element of our ring $R$ is assigned to each $k$-cell in a cell complex $K$, creating a function: $$\tau _k: Q_k(K) \rightarrow R.$$ It is extended by linearity to the whole chain complex of the cell complex: $$\tau _k: C_k(K) \rightarrow R.$$ These functions are called cochains, or differential forms.

Now, in any chain complex $(C,\partial)$ we can define cochains as linear maps to the coefficient ring: $$\tau _k: C_k \rightarrow R.$$ Then we recognize that they form the dual space $(C_k)^*$ of $C_k$ over $R$. It is denoted by $C^k$.

The dual of the boundary operator $$\partial _k^C:C_k \rightarrow C_{k-1}$$ is the coboundary operator: $$\partial ^k_C:C^k \leftarrow C^{k-1}.$$

The combination of all of these spaces with the coboundary operators is denoted by $(C^*,\partial ^*_C)$ and is called the cochain complex of $C$. See discrete exterior calculus.

We don't separate "quantities" from "transformations". We just think of cochains on $C$ as linear map that constitute a chain map from $C$ to certain trivial chain complex: $$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} \newcommand{\la}[1]{\!\!\!\!\!\!\!\xleftarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\ua}[1]{\left\uparrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{ccccccc} \da{\partial_3} & \searrow ^{df_2} & \da{\partial_3} \\ C_2 & \ra{f_2} & R\\ \da{\partial_2} & \searrow ^{df_1} & \da{\partial_2} \\ C_{1} & \ra{f_{1}} & R\\ \da{\partial_1} & \searrow ^{df_0} & \da{\partial_1} \\ C_{0} & \ra{f_{0}} & R\\ \da{\partial_0} & & \da{\partial_0 =0} \\ 0& &0\\ \end{array} $$ The diagonal arrows show the exterior derivatives of these forms following the Stokes Theorem.

For a given $k$-cochain $f=f_k$ we can construct the rest of such a chain map: $$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} \newcommand{\la}[1]{\!\!\!\!\!\!\!\xleftarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\ua}[1]{\left\uparrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{ccccccc} \da{\partial_{k+2}} & & \da{\partial_{k+2} =0} \\ C_{k+2} & \ra{f_{k+2}=0} & R\\ \da{\partial_{k+1}} & & \da{\partial_{k+1} =0} \\ C_{k+1} & \ra{f_{k+1}=f_k\partial _k} & R\\ \da{\partial_{k}} & & \da{\partial_k=Id} \\ C_{k} & \ra{f_{k}} & R\\ \da{\partial_{k-1}} & & \da{\partial_{k-1} =0} \\ C_{k-1}& \ra{f_{k-1}=0} & R\\ \end{array} $$ We choose $f_{k+1}=f_k\partial _k$ to make the lower square commute. We also choose $f_{k+2} =0$. Then the upper square commutes because, since $C$ is a chain complex, we have $$f_{k+1}\partial _{k+1}=f_k\partial _k \partial _{k+1}=f_k 0 = 0.$$ (Cf. chain map over a natural transformation.)

Note that it is easy to find a cell complex with this chain complex, see Homology of balls and spheres.

So, in a sense, $$(C^*,\partial ^*) = \text{Hom}(C,R) \in \text{Hom}(Ch(\mathcal{Mod})).$$

Meanwhile, the dual of a chain map $$f \colon ( S, \partial ^S) \rightarrow (T, \partial ^T)$$ is a cochain map, i.e., a linear operator between cochain complexes $$f^* \colon ( S^*, \partial ^*_S) \leftarrow (T^*, \partial ^*_T)$$ that preserves coboundaries. It is a sequence of linear operators $$f^*=\{ f^k \colon S^k \leftarrow T^k\}$$ that commute with the duals of the boundary operators: $$ \partial_S^k f^{k-1} = f^k \partial_T ^k,$$ or simply: $$\partial ^* f^* = f^* \partial ^* .$$

Then the arrow in the above diagram are reversed and the indices raised: $$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} \newcommand{\la}[1]{\!\!\!\!\!\!\!\xleftarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\ua}[1]{\left\uparrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{ccccccc} \ua{\partial^3_K} & & \ua{\partial^3_L} \\ C^2(K) & \la{f^2} & C^2(L) \\ \ua{\partial^2_K} & & \ua{\partial^2_L} \\ C^{1}(K) & \la{f^{1}} & C^{1}(L)\\ \ua{\partial^1_K} & & \ua{\partial^1_L} \\ C^{0}(K) & \la{f^{0}} & C^{0}(L)\\ \ua{\partial^0_K} & & \ua{\partial^0_L} \\ 0& &0\\ \end{array} $$