I'm currently reading the sketch proof of Theorem 1 in Beavuille's notes on the Luroth problem. Let $P$ be the blow-up of $\mathbb{P}^3$ in finitely many points and smooth curves, and let $X \subset \mathbb{P}^4$ be a smooth cubic threefold.
In the proof, it says that since we have constructed a birational morphism $f : P \to X$, we have the homomorphisms of cohomology groups $f_* : H^3(P, \mathbb{Z}) \to H^3(X, \mathbb{Z})$ and $f^* : H^3(X, \mathbb{Z}) \to H^3(P, \mathbb{Z})$ with $f_* f^* = 1$, compatible with the Hodge decomposition and cup products. Thus, $H^3(X, \mathbb{Z})$ with its polarised Hodge structure taken into account is a direct summand of $H^3(P, \mathbb{Z})$.
Q1: Why does the beginning of the above paragraph imply that $H^3(X, \mathbb{Z})$ is a direct summand of $H^3(P, \mathbb{Z})$? Also, why not the other way round?
The proof then goes on to say that the above fact about cohomology groups implies that $J(X)$ (where $J$ means intermediate Jacobian) is a direct summand of $J(P)$.
Q2: Why does this follow? In our cubic threefold case, $J(X)$ is defined as $$ J(X) = \frac{H^{1,2}(X)}{H^3(X, \mathbb{Z})_{\mathrm{tf}}} \cong \frac{H^2(X, \Omega_X)}{H^3(X, \mathbb{Z})_{\mathrm{tf}}} $$ where the subscript $\mathrm{tf}$ means torsion-free part of the cohomology group. So let's say as we deduced above that $H^3(X, \mathbb{Z})$ is a direct factor of $H^3(P, \mathbb{Z})$. So then $H^3(P, \mathbb{Z}) \cong G \oplus H^3(X, \mathbb{Z})$ for some group $G$. Then $$ J(P) = \frac{H^{1,2}(P)}{H^3(P, \mathbb{Z})_{\mathrm{tf}}} \cong \frac{H^{1,2}(P)}{H^3(X, \mathbb{Z})_{\mathrm{tf}}\oplus G} $$ but I don't see how we can say from this that $J(X)$ must be a factor of $J(P)$? I suppose if we also had $H^{1,2}(P)$ splitting as $H^{1,2}(X)$ plus something else, I'd believe this - but I can't see why this would be true? I guess there is a short exact sequence relating the cotangent sheaf of a blow-up to the thing being blown up, but $f : P \to X$ isn't a blow-up so this doesn't seem useful.
And another question:
Q3: Why does Beauville take the torsion free part of the integral cohomology in the definition of $J(X)$, whereas in e.g. Voisin's book, the definition of $J(X)$ seems to just have normal integral cohomology?
Thanks.