Let $f:X \to Y$ be a proper surjective morphism of reduced connected noetherian schemes. Assume $Y$ is irreducible. Let $y \in Y$ be a closed point. Denote by $X\_y$ the fiber over $y$ to the morphism $f$. Suppose that $X\_y$ is a non-singular, irreducible variety and for every closed point $x \in X\_y$, the tangent space $T\_xX$ have the same dimension. Does this imply that any irreducible component of $X$ intersecting $X\_y$ contains the whole of $X\_y$? Can we drop the assumption of reducedness of $X$ or $Y$?



Relative tangent space to proper morphism and irreducibility of fibers
That is certainly not true. Let $Y$ be $\mathbb{A}^2$ with coordinates $(s,t)$. Denote by $[u\_0,u\_1]$ homogeneous coordinates on $\mathbb{P}^1$. Let $X$ be the hypersurface of $\mathbb{A}^2\times \mathbb{P}^1$ with homogeneous defining equation $stu\_1 = 0$. Let $f:X\to Y$ be the restriction to $X$ of the projection to $Y$. Let $y$ be the closed point with associated maximal ideal $\langle s,t \rangle$. 


The fiber $X\_y$ of $f$ over $y$ is $\mathbb{P}^1$. The Zariski tangent space to $X$ at every point of $X\_y$ equals the entire Zariski tangent space to $\mathbb{A}^2\times \mathbb{P}^1$, and this is $3$-dimensional. The irreducible component where $u\_1$ equals $0$ intersects $X\_y$, but it does not contain $X\_y$.


There is a similar example where $Y$ is an irreducible nodal curve and $y$ is the node. 


