The set up is $C$ is a curve and $J$ is its Jacobian. On the $C \times J$ there is the Poincare bundle $P$ which is the universal family of degree zero line bundles on $C$. For every integer $d$ there is also a line bundle $P(d)$ on $C \times J$ which is a family of line bundles of degree $d$ on $C$.

I've seen a construction of $P$ and given $P$ an example of a $P(d)$ would be $P^L:= q\_1^\*L\otimes P$ where $q\_1 \colon C \times J \to C$ is the projection and $L$ is a line bundle of degree $d$ on $C$. If $L,L'$ both have degree $d$ you can form either $P^L$ or $P^{L'}$. I've seen $P(d)$ defined as the inverse limit of $P^L$ as $L$ ranges over all degree $d$ line bundles. I don't really know how to think of such an inverse limit. Is there a more concrete way to describe $P(d)$? It seems up to isomorphism, $P(d) = P^L$, but this is "very" non canonical which seems bad. For example does $P^L$ have a universal property like $P$ does?

I have a more specific question related to this. This question is coming from Prop. 21.6 of Polishchuk's book on Abelian varieties and the Fourier Mukai tranform, if anyone is curious. If $q\_2 \colon C \times J \to J$ is the projection, then apparently $q\_{2\*}P(g-1) = 0$. Why is this so?

One argument for this (that I don't understand) is the following.

1) Embed $P(g-1) \to F$ with $F$ flat over $J$ and $R^1q\_{2\*}F = 0$.

2) From $0 \to P(g-1) \to F \to F/P=: E \to 0$ and cohomology we get

$0 \to q\_{2\*}P(g-1) \to q\_{2\*}F \to q\_{2\*}F \to R^1q\_{2\*}P(g-1) \to 0$

3)The restriction of $R^1q\_{2\*}P(g-1)$ to $a \in J$ is $H^1(C \times a, P(g-1)|\_{C \times a})$ which is $H^1$ of a line bundle of degree $g-1$. This is zero outside the theta divisor and Riemann-Roch says $h^1(L) = h^0(L)$ for degree $g-1$ line bundles, so 

$q\_{2\*}F \to q\_{2\*}F$

is an isomorphism outside a divisor. And hence

4) $q\_{2\*}F \to q\_{2\*}F$ is injective.

Part 4) is the part I don't see.

I don't necessarily want to understand this line of reasoning. But I don't see why $q\_{2\*}P(g-1) = 0$ since it seems to be supported on the theta divisor.