# Cohomology with $\mathbb{Z}_2$ coefficients, wedge product and integrals

I'm studing cohomology with $\mathbb{Z}_2$ coefficients in order to understand Stiefel-Whitney classes, and i have a couple of questions.Let $X$ be a n-manifold.

1) I know that $H^k_{sing}(X,\mathbb{Z})\subset H^k_{sing}(X,\mathbb{R})\simeq H^k_{DR}(X)$, and that the wedge product $\wedge$ is the cup product $\cup$ for $H^k_{DR}(X)$. Thank to the mentioned inclusion, i can use the wedge product also on $H^k_{sing}(X,\mathbb{Z})$. Now i am wondering what happens for singular cohomology with $\mathbb{Z_2}$ coefficients. A priori no wedge product is defined, but using the Universal coefficient theorem, if $Ext (H_{k-1}(X,\mathbb{Z}),\mathbb{Z}_2)=0$ then every element in $H^k(X,\mathbb{Z_2})$ is the reduction mod 2 of an element in $H^k(X,\mathbb{Z})$, but on $H^k(X,\mathbb{Z})$ the wedge product is defined, while on $H^k(X,\mathbb{Z_2})$ is not. how does it work? when i reduce mod 2 the wedge product becomes the cup product?

2) Given $\omega \in H^k(X,\mathbb{R})$ and $[S]$ a k-dimensional cycle i know we have the pairing $\langle [S],\omega\rangle:= \int_s\omega$. How is this defined for $H^k(X,\mathbb{Z}_2)$? Apparently, when $\omega \in H^k(X,\mathbb{Z}_2)$ is the reduction mod 2 of $\widetilde{w}\in H^k(X,\mathbb{Z})$, it is ok to set $\langle [S],\omega\rangle=\int_S \widetilde{\omega}$ mod 2. For example in the proof of the wu's formula http://books.google.it/books?id=nqMNrHE1U28C&pg=PA163&lpg=PA163&dq=%22wu's+formula%22&source=bl&ots=ksKOp7Hw6C&sig=8GQt7P48siLl8YGnqSI4HRQjckk&hl=it&sa=X&ei=DrtAUeD2KoSq4AS4vYCIBg&ved=0CC8Q6AEwAA#v=onepage&q=%22wu's%20formula%22&f=false (here the notation is $w_2(T_M)\cdot S$) Scorpan says that $w_2(T_S)$is the reduction mod 2 of the euler class, but the integral must be implicit because $\int_S e(T_S)=\mathcal{X}(S)$, but is the integral defined for elements in $H^k(X,\mathbb{Z_2})$? How does it work?

Thank you

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Your inclusion $H^k_{sing}(X,\mathbb Z) \subset H^k_{sing}(X,\mathbb R)$ is not true. Cohomology with integer coefficients can have torsion. The inclusion is missing a tensor product with $\mathbb R$. Or perhaps you want to think about the free part (mod torsion) for cohomology with integer coefficients. –  Ryan Budney Mar 13 '13 at 18:16
yes, i'm sorry, i meant that, integer coefficients cohomology modulo torsion –  Konrad Mar 13 '13 at 18:38
The pairing between cohomology and homology uses only the definition of cohomology, as co-cycles modulo co-boundaries. A co-cycle is a homomorphism from the chain group to the integers (or whatever coefficients you choose). So you can always evaluate such a cochain on a chain. Your thread has some votes to close as "too localized". I think that is because part of your question seems to indicate lack of familiarity with the definition of cohomology. –  Ryan Budney Mar 13 '13 at 22:50
Why not pick up a book like Bredon's "Geometry and Topology" and work through the homology/cohomology section, specifically the universal coefficient theorem. This would answer all your questions, as far as I can tell. –  Ryan Budney Mar 13 '13 at 22:51