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Given a morphism between two smooth toric varieties $f: X \rightarrow Y$, is the dimension of the kernel of $\mathrm{d}f$ at any point $p \in X$ equal to the dimension of the fiber at $f(p) \in Y$?

I think this is true when $Y = \mathbb{P}^n$ and $X$ is obtained from $Y$ by a finite sequence of blow-ups (by local coordinates calculations), but I am not sure how general this statement is. If this is true, how to prove it? If this is not true, could someone tell me a counter-example? Thank you.

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    $\begingroup$ Does a ramified double cover of $\mathbb{P}^1$ by itself give a counterexample at the ramification locus? $\endgroup$
    – S. Carnahan ♦
    Commented Jun 25, 2014 at 17:33
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    $\begingroup$ In fact, even if the morphism is torus equivariant, and it is an isomorphism on the open orbit, this can still fail. Just consider the iterated blowup of the affine plane, first at the origin, then at one of the two torus invariant points on the exceptional divisor. The fiber of this new scheme over the origin in the plane is a union of two $\mathbb{P}^1$s, and the condition fails at the intersection point. $\endgroup$
    – Jason Starr
    Commented Jun 25, 2014 at 19:13

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