# Cap product

In algebraic topology the cap product is a method of adjoining a chain of degree p with a cochain of degree q, such that qp, to form a composite chain of degree pq. It was introduced by Eduard Čech in 1936, and independently by Hassler Whitney in 1938.

## Definition

Let X be a topological space and R a coefficient ring. The cap product is a bilinear map on singular homology and cohomology

${\displaystyle \frown \;:H_{p}(X;R)\times H^{q}(X;R)\rightarrow H_{p-q}(X;R).}$

defined by contracting a singular chain ${\displaystyle \sigma :\Delta \ ^{p}\rightarrow \ X}$ with a singular cochain ${\displaystyle \psi \in C^{q}(X;R),}$ by the formula :

${\displaystyle \sigma \frown \psi =\psi (\sigma |_{[v_{0},\ldots ,v_{q}]})\sigma |_{[v_{q},\ldots ,v_{p}]}.}$

Here, the notation ${\displaystyle \sigma |_{[v_{0},\ldots ,v_{q}]}}$ indicates the restriction of the simplicial map ${\displaystyle \sigma }$ to its face spanned by the vectors of the base, see Simplex.

## Interpretation

In analogy with the interpretation of the cup product in terms of the Künneth formula, we can explain the existence of the cap product by considering the composition

${\displaystyle C_{\bullet }(X)\otimes C^{\bullet }(X){\overset {\Delta _{*}\otimes \mathrm {Id} }{\longrightarrow }}C_{\bullet }(X)\otimes C_{\bullet }(X)\otimes C^{\bullet }(X){\overset {\mathrm {Id} \otimes \varepsilon }{\longrightarrow }}C_{\bullet }(X)}$

in terms of the chain and cochain complexes of ${\displaystyle X}$, where we are taking tensor products of chain complexes, ${\displaystyle \Delta \colon X\to X\times X}$ is the diagonal map which induces the map ${\displaystyle \Delta _{*}}$ on the chain complex, and ${\displaystyle \varepsilon \colon C_{p}(X)\otimes C^{q}(X)\to \mathbb {Z} }$ is the evaluation map (always 0 except for ${\displaystyle p=q}$).

This composition then passes to the quotient to define the cap product ${\displaystyle \frown \colon H_{\bullet }(X)\times H^{\bullet }(X)\to H_{\bullet }(X)}$, and looking carefully at the above composition shows that it indeed takes the form of maps ${\displaystyle \frown \colon H_{p}(X)\times H^{q}(X)\to H_{p-q}(X)}$, which is always zero for ${\displaystyle p.

## The slant product

The above discussion indicates that the same operation can be defined on cartesian products ${\displaystyle X\times Y}$ yielding a product

${\displaystyle \backslash \;:H_{p}(X;R)\otimes H^{q}(X\times Y;R)\rightarrow H^{q-p}(Y;R).}$

In case X = Y, the two products are related by the diagonal map.

## Equations

The boundary of a cap product is given by :

${\displaystyle \partial (\sigma \frown \psi )=(-1)^{q}(\partial \sigma \frown \psi -\sigma \frown \delta \psi ).}$

Given a map f the induced maps satisfy :

${\displaystyle f_{*}(\sigma )\frown \psi =f_{*}(\sigma \frown f^{*}(\psi )).}$

The cap and cup product are related by :

${\displaystyle \psi (\sigma \frown \varphi )=(\varphi \smile \psi )(\sigma )}$

where

${\displaystyle \sigma :\Delta ^{p+q}\rightarrow X}$ , ${\displaystyle \psi \in C^{q}(X;R)}$ and ${\displaystyle \varphi \in C^{p}(X;R).}$

An interesting consequence of the last equation is that it makes ${\displaystyle H_{\ast }(X;R)}$ into a right ${\displaystyle H^{\ast }(X;R)-}$ module.