The parallels between the formulas in Schubert calculus and in the theory of the representations of symmetric groups (par GeissingerZelevinsky) are so apparent (e.g. Giambelli formula), that one must wonder how to directly define the comultiplication on Schubert (co)cells.
1 Answer
If you're talking about Grassmannians, then you can use the direct sum maps
$Grass(k,n) \times Grass(k',n') \to Grass(k+k', n+n')$
to get the comultiplication. This induces a map on cohomology the other way. Then take the limit as $n,n' \to \infty$ and then take the limit $k,k' \to \infty$. This fixes two problems: 1) the Grassmannians aren't the same in the finite case, and 2) the values of $k,n$, etc. give truncations of the ring of symmetric functions, so you need to remove that restriction.
According to Symmetric polynoms are Hopf algebra ? What for one needs coproduct ? the bialgebra is enough to get the whole Hopf structure.
Positivity (to get the PSH algebra structure) follows from geometric considerations (i.e., all structure coefficients are intersection numbers).

$\begingroup$ Steve  thanks! Actually, I wanted something more elementary, in terms of intersections. And from your answer it is clear how to obtain it: in Schubert calculus multiplication is intersection, comultiplication is the inverse image of $Grass(\infinity, \infinity) \times Grass(\infinity, \infinity) \to (\infinity, \infinity)$. Apparently, inverse image commutes with the intersection. $\endgroup$– GeorgeJun 6, 2013 at 2:27

$\begingroup$ Another attempt to get a nice formula: $Grass(\infty,\infty) \times Grass(\infty,\infty) \to Grass(\infty, \infty)$ $\endgroup$– GeorgeJun 6, 2013 at 2:32