Tangent bundles and birational morphisms

Let $f:X\rightarrow Y$ be a smooth morphism between smooth schemes. Then there is an exact sequence $$0\mapsto T_{X/Y}\rightarrow T_{X}\rightarrow f^{*}T_{Y}\mapsto 0$$ Now let us assume $f$ to be a birational morphism with smooth fibers, meaning that the exceptional locus of $f$ is the disjoint union of smooth divisors. In this setting does the above exact sequence still exist?

• The sequence exists, but it won't be exact on the right. You always have the map $df:T_X\to f^* T_Y$, and you define $T_{X/Y}$ to be the kernel, but of course $df$ will not be surjective where $f$ is not smooth (in fact, smoothness of $f$ is equivalent to surjectivity). – Piotr Achinger Oct 5 '13 at 1:33
• And moreover, the kernel of $df$ is zero in this case. – Sasha Oct 5 '13 at 3:28

Unless the morhism $f$ is etale, this exact sequence is not true in general. The problem is this: the differantial $df$ can be viewed as an injection $f^{*}K_{Y}\to \Omega^{1}_{X}$. One defines $\Omega^{1}_{X/Y}$ to be the cokernel of this injection, so that we have an exact sequence: $0\to f^{*}K_{Y}\to \Omega^{1}_{X} \to \Omega^{1}_{X/Y}\to 0$. So far is true, but when you dualize, you do not get the exact sequence $0\to T_{X/Y}\to T_{X}\to f^{*}T_{Y}\to 0$. Because $T_{X}\to f^{*}T_{Y}$ is not surjective in general. There is an $Ext^{1}$ which comes in after dualizing which is the cokernel of this non-surjective map. So as Piotr pointed out, there will only exist an exact sequence $0\to T_{X/Y}\to T_{X}\to f^{*}T_{Y}$.