I've gone through many texts in algebraic geometry, specifically, Schubert calculus. They all claim that the Schubert classes $[\Omega\_{\lambda}]$ form a basis for the cohomology ring of the complex Grassmannian of $k$-dimensional subspaces of $n$-dimensional complex space, i.e. $H^{\*}(Gr(k, n))$, but without any proof. I would think that there could be a way to somehow show this result using an argument involving linear independence and span, but I'm not sure where to begin. In the back of Fulton's $\textit{Young tableaux}$, it is suggested to construct a filtration $$Gr(k, n)=Y\_0 \supset Y\_1 \supset \cdots,$$ where $Y\_p$ is the union of all Schubert varieties $\Omega\_{\lambda}$ with $|\lambda| \geq p$. Then I would have to take the cohomology of each member of the filtration, but how does one conclude that the Schubert classes form a basis of the cohomology ring in question?



How do Schubert classes form a basis for $H^{*}(Gr(k, n))$?
What Fulton is suggesting is (an easy special case of) the fact that cellular and usual cohomology are equal. Consider the long exact sequence attached to the pair $(Y\_p,Y\_{p-1})$. We have maps
$$ \cdots \to H^\*(Y\_{p}, Y\_{p-1}; \mathbb Z)\to H^\*(Y\_p; \mathbb Z) \to H^\*(Y\_{p-1}; \mathbb Z) \to \cdots.$$


The key observation is that $Y\_{p}$ with $Y\_{p-1}$ smashed to a point is a wedge of $2p$ dimensional spheres, one for each Schubert cell of dimension $p$. Thus $H^i(Y\_{p}, Y\_{p-1}; \mathbb Z)$ is only non-zero for $i=2p$, and that degree is the free abelian group on the Schubert classes in dimension $p$. Now, induct on dimension, and the result follows.


