I apologize if this question is too basic, but I haven't been able to work this out for myself.


Let $X$ and $Y$ be projective schemes, say over the complex numbers. There exists a scheme $Hom(X,Y)$ parameterizing morphisms $f : X \to Y$. In Kollar's "Rational curves on algebraic varieties" it is proved that the stalk of the tangent sheaf of $Hom(X,Y)$ at a point $[f]$ is
$$
T\_{Hom(X,Y),[f]} = H^0(X, Hom(f^\*\Omega^1\_Y, \mathcal O\_X)).
$$
I suppose that if $X$ and $Y$ are sufficiently nice, this coincides with $H^0(X,f^\*T\_Y)$, but that's not important for now.


Can we obtain the tangent sheaf of $Hom(X,Y)$ "globally"? What I mean is, suppose we consider the evaluation morphism $ev : X \times Hom(X,Y) \to Y$ given by $(x,f) \mapsto f(x)$ and the projection $p : X \times Hom(X,Y) \to Hom(X,Y)$. The morphism $p$ is proper since $X$ is compact. Then the sheaf 
$$
\mathcal T := p\_{\ast} Hom({ev^\ast} \Omega^1\_Y,\mathcal O\_X)
$$ 
over $Hom(X,Y)$ is coherent, since it is the direct image of a coherent sheaf on the product space. It also has the same stalks as the tangent sheaf of $Hom(X,Y)$. Is $\mathcal T$ the tangent sheaf of $Hom(X,Y)$?



Tangent sheaf of a hom scheme
There are references for the statement with the sheaf of relative differentials. Although it is not the original reference, one reference is on p. 83, the discussion immediately preceding Proposition 6.3 of the following article.


MR1437495 (98e:14022) Reviewed
Behrend, K.(3-BC); Fantechi, B.(I-TRNT)
The intrinsic normal cone. (English summary)
Invent. Math. 128 (1997), no. 1, 45–88.
14F99 (14C15 14D20)


