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 $Hom(X,Y)$?