Is multiplication by a Stiefel-Whitney class an injective map?

Question

I have a doubt: In cohomology, when you multiply by a Stiefel-Whitney class is it always an injective map? For example: is $$H^{j-1}(X)\xrightarrow{\smile\ w_1}H^{j}(X)$$ always injective?

Thanks!

Answer 1

As Qiaochu Yuan pointed out, there is no reason to expect this to be the case, and it is not true in the situation he described. Here are some other examples where taking the cup product with the first Stiefel-Whitney class is not injective.

Consider $X = S^1\times S^1$. Then $w_1(X) = 0$ as $X$ is orientable, so the map

$$H^1(X; \mathbb{Z}_2) \xrightarrow{\cup\ w_1(X)} H^2(X; \mathbb{Z}_2)$$

is not injective as it is the zero map and $H^1(X; \mathbb{Z}_2) \neq 0$.

Even if the first Stiefel-Whitney class is non-zero, the map you describe may not be injective.

Consider $Y = \mathbb{RP}^2\times\mathbb{RP}^2$. As $\mathbb{RP}^2$ is not orientable, $w_1(\mathbb{RP}^2) \neq 0$; let $a = \pi_1^*w_1(\mathbb{RP}^2)$ and $b = \pi_2^*w_1(\mathbb{RP}^2)$ where $\pi_1, \pi_2 : \mathbb{RP}^2\times\mathbb{RP}^2 \to \mathbb{RP}^2$ are the projections onto the first and second factors respectively. Then $H^*(Y; \mathbb{Z}_2) \cong \mathbb{Z}_2[a, b]/\langle a^2, b^2\rangle$ and $w_1(Y) = a + b$. The linear map

$$H^1(Y; \mathbb{Z}_2) \xrightarrow{\cup\ w_1(Y)} H^2(Y; \mathbb{Z}_2)$$

is not injective. To see this, note that $a + b$ is a non-zero element of $H^1(Y; \mathbb{Z}_2)$ but

$$(a + b)\cup w_1(Y) = (a + b)\cup(a + b) = a^2 + a\cup b + b\cup a + b^2 = 2(a\cup b) = 0.$$

Answer 2

The Stiefel-Whitney class $w_1$ can be arbitrary, so you're asking whether multiplication by some class $w_1 \in H^1(X, \mathbb{Z}_2)$ is always injective. The answer is of course no: if $X$ is, say, a manifold, its cohomology vanishes above some dimension, so at some point multiplication by any cohomology class necessarily fails to be injective.