# Bockstein Homomorphism and real projective space.

algebraic-topologygeneral-topologygeometric-topologysolution-verification

Consider the Bockstein Homomorphism $$\beta: H^1(\mathbb{R{P}}^2, \mathbb{Z}_2)\rightarrow H^2(\mathbb{R}P^2,\mathbb{Z}_2)$$. Now, I was told that

For each $$\alpha$$, $$\beta(\alpha) = \alpha \cup \alpha$$. I am trying to prove this. My idea is that $$\alpha =0$$ then $$\alpha \cup \alpha=0$$ so all we are left to do is establish that for $$\alpha$$; generator of $$H^1(\mathbb{R}P^2,\mathbb{Z}_2)=\mathbb{Z}_2$$, $$\alpha \cup \alpha$$ is a generator. This is because $$\beta$$ is non trivial and must therefore map generator to generator. To do so, all I need to do is use poincare duality, and then note that $$\alpha$$ is the dual of $$\alpha \cap [\mathbb{R}P^2]_{\mathbb{Z}_2}$$ and thus must evaluate to $$1$$.

Is this approach correct?

This is an instance of the general fact that the Steenrod operation $$\operatorname{Sq}^i$$ squares elements of degree $$i$$: $$\operatorname{Sq}^i(a) = a \cup a$$ for $$|a|=i.$$ It is known that $$\operatorname{Sq}^1 = \beta$$.
Now, if you want to see it directly, then yes: first of all, you note that $$\beta$$ maps the generator to the generator: this is seen directly from the definition of $$\beta$$ by writing chain complexes and tracing the image of the generator $$a$$ of $$H^1.$$ Then knowing that the cup product $$a \cup a$$ if nontrivial will mean that it is equal to $$\beta(a).$$ Nontrivialoty of the cup product on the projective space is a classical (and pivotal) result, and there are many ways to prove it. The one you suggest is one of the least painful. You note that cup product corresponds via Poicare duality to intersecting things and use the geometry of $$\mathbb{R}P^2$$ to deduce the result.