Let $S$ be a smooth projective surface (I am mostly intrested in the case when $S$ is a product of curves, say $S=\mathbb{P}^1 \times \mathbb{P}^1$ but probably this is not important). 


Consider a family of curves $X \subset S \times T$ parametrised by a variety $T$ of dimension 2 (the fibres $X\_t$ are distinct). Denote the projection $X \to S\ $ as $p\_S$ and $X \to T\ $ as $p\_T$. Define a map $\tau: X \to \mathbb{P}TS$ which to every point $x \in X$ associates the projectivisation of the tangent vector of $X\_{p\_T(x)}$ at $p\_S(x)$. 


My question is: how can one show that there exists $s \in S$ such that the map $\tau$ is non-constant on $p^{-1}(s)$?


In other words, how can one show that there is a point $s \in S$ such that not all curves passing through it touch each other, but on the contrary, their tangent spaces sweep the tangent space of $S$ at this point?


**update**: Consider the map $\sigma: \mathbb{P}(TX/T) \to \mathbb{P}(TS)$ induced by the projection $p\_S: X \to S$. If $\sigma^{-1}(v)$ were finite for some $v \in TS$, then the projection of $v$ to $S$ would be the answer to the question. If the image of the map $\sigma$ is of dimension 3 then almost all fibres are of dimension 0, since $\mathbb{P}(TX/T)$ is of dimension 3. Perhaps one can show that $\mathrm{dim}\ \mathrm{Im}\ \sigma=3$? 



base points of multiplicity $>1$
I guess I post an answer to close the issue.


The statement is true in characteristic 0, but not in positive characteristic. Conterexamples in the latter case are easy to come by an $S=\mathbb{A}^1 \times \mathbb{A}^1= \mathbb{A}^2$, say, using curves with projections on one of the $\mathbb{A}^1$ factors everywhere ramified.


In characteristic 0 the argument is as follows, working over $\mathbb C$ (one would have to use formal completions of local rings and formal power series, put on the level of ideas the proof is the same).


So suppose the statement is not true. Then one can find a point $y$ and a neighbourgood $z$ such that for any $z \in U$ and any curve $X\_t$ incident to $Q'$ the image of $T\_x X\_t$ in $T\_y S$ depends only on $y$. In other words all curves $X\_t$ are integral with respect to a distribution on $S$. But by uniqueness of solutions of ordinary differential equations, there can be at most one integral curve incident to each $z \in U$, which contradicts that dimension of $T$ is 2.


