products and smooth/étale/unramified morphisms Let $X$, $Y$ and $Z$ be Noetherian schemes. 
If $f: Y \to X$ and $g: Z \to X$ are morphisms of finite type, such that at each point of $X$, at least one of the two morphisms is smooth/étale/unramified (at all points of its inverse image), can we conclude that the induced morphism $Y \times_X Z \to X$ is smooth/étale/unramified everywhere? 
If not, which results can we obtain?
(In his textbook on Algebraic Geometry, Liu asks to prove that the answer is always "yes"...)
EDIT. So, indeed, the problem statement in the book is wrong...
 A: No.  As an extreme example, suppose that $g$ is the identity (which is etale everywhere), and that $f$ is not etale at some point.  Then the fibre product is just $f$ again.
But in fact, this is essentially the general case.  If $g$ is etale (or smooth) at a point, then it is etale (resp. smooth) in a n.h. of that point, so we may replace $Z$ by the n.h. and so assume that
$g$ is etale everywhere.  Then if $f$ is not etale (or smooth) at a point $y \in Y$, the product will not be etale in a n.h. of $y \times Z.$
(Imagine that $Y$ was e.g. a nodal curve with a node at $y$, and that $Z$ is a smooth
curve.  (Here $X$ is Spec of the ground field.)  Then $Y\times Z$ is the product of a nodal
curve and a smooth curve, which just looks like a cylinder over the nodal curve; it is 
singular all along the "cylinder" over the node.)
A: A correct statement would be : suppose Y is unramified/étale/smooth over X, then $Y\times_X Z\to X$ is unramfied/étale/smooth iff $Z\to X$ is. 
[EDITED] According to Matt's remark below, we must suppose $Y\to X$ surjective. 
A: Smooth, unramified, and etale morphisms are stable under base change, so $Y\times_X Z \to Z$ is {smooth, unramified, etale} if $Y\to X$ is {smooth, unramified, etale}.
