Let $X$ and $Y$ be smooth algebraic varieties over a field $k$ of characteristic $0$. For varieties we know that $X/k$ is rigid if and only if $H^{1}(X,T\_{X})=0$. But $H^{1}(X,T\_{X})$ also parametrizes the first order deformations. So in fact we have that there is no infinitesimal deformations if there is no first order infinitesimal deformation. My question is that does the same hold true for deformations of morphisms? Namely, let $f\_{0}:X\to Y$ be a morphism. Then we know that first order deformations are parametrized by the space $H^{0}(X,f\_{0}^{\*}T\_{Y})$ ($T\_{Y}$, tangent bundle of $Y$). Is it true that $f\_{0}$ is rigid if $H^{0}(X,f\_{0}^{\*}T\_{Y})=0$? i.e. if there is no non-trivial first order deformation? If there is a reference for this in the literature, I'd be grateful to know. 



Characterizing the rigidity of morphisms of smooth varieties
The space $H^{0}(X,f\_{0}^{\*}T\_{Y})$ is the tangent space at $f\_0$ to the variety $\mathrm{Hom}(X,Y)$, see J. Kollár, *Rational curves on algebraic varieties*. Thus if it is zero, $f\_0$ is an isolated point, hence is rigid. However I don't think the converse holds, even for deformations of $X$: $X$ may be rigid with $H^1(X,T\_X)\neq 0$. I don't have any example at hand but I am pretty sure that they exist. 


