Restriction of a dominant map Let $X$ be a variety over a field $k$, and suppose that we have a dominant map $\varphi: \mathbb{A}^{m} \dashrightarrow X$. Assume that $m > \dim(X)$. Then we can find a dense open set $U\subset \mathbb{A}^{m}$ such that $\varphi|_{U}$ is an open map whose fibers are $m-\dim(X)$ dimensional. Let $u\in U$ be a point. I would like to understand the next statement:

If $u \in Z \subset \mathbb{A}^{m}$ is a hypersurface which does not
  contain the irreducible component of the fiber of $\varphi|_{U}$
  through $u$, then $\varphi|_{Z}: Z \dashrightarrow X$ is dominant.

What is the geometric intuition here? Why does restricting a dominant map to a certain hypersurface preserve the dominance? 
 A: Here's an example that easy to visualize: let $m=2$ and let $X=\mathbb A^1$ and let the map be the projection onto the $x$ axis. The fibers are $1$-dimensional lines parallell with the $y$ axis (so in this case we can take $U=\mathbb A^2$).
Now the point is that unless the hypersurface is contained in one of the fibers (that is, one of the lines parallell with the $y$-axis), then the map is dominant. Note that in this case, a hypersurface is just a curve in $\mathbb A^2$.
A: A good example is$$\mathbb{A}^3 \to \mathbb{A}^2,\text{ }(x,y,z) \mapsto (y,z),$$and now consider the hypersurface $Z = V(yz)$ and the point $u=(0,0,0)$. The issue is that $Z$ maps onto $V(yz)$ which is a proper subvariety of $\mathbb{A}^2$, and notice that $Z$ contains the fiber $(x,0,0)$ over $(0,0)$.
For the general problem, it helps to now look at the algebra---a morphism of affine varieties is dominant if the pullback on coordinate rings is injective.

Exercise (The pullback map between coordinate rings). Suppose that $F: X \to Y$ is a morphism of affine varieties.

*

*Show that $F^*: k[Y] \to k[X]$ is injective if and only if $F$ is dominant, i.e. the image set $F(X)$ is dense in $Y$.

*Show that $F^*: k[Y] \to k[X]$ is surjective if and only if $F$ defines an isomorphism between $X$ and some algebraic subvariety of $Y$.

*Find an example where $F$ is injective but $F^*: k[Y] \to k[X]$ is not surjective.


And we have$$k[Y] \to k[X] \to k[Z]=k[X]/(f)$$where $Z=V(f)$ say. The first map $\phi^*$ is injective by the dominance assumption on $\phi$, and for $Z \to Y$ to be dominant we need the above composite to be injective.
In the example, the issue is that injectivity fails as$$yz \mapsto yz \mapsto 0,$$the key being that $yz$ in $k[X]=k[x,y,z]$ has a (unique) preimage in $k[Y]=k[y,z]$.
In general, the issue is that $f$ might lie in the image of $k[Y] \to k[X]$ (or more generally, a $k$-linear combination of positive powers of $f$ may be in the image).
We now use the assumption that $u$ lies in $Z$, algebraically---$f$ lies in the maximal ideal $\mathfrak{m}_u$ of $k[X]$ (abbreviating by $\mathfrak{m}_p$ the functions that vanish at a point $p$).
Now, I think one can check that $f$ lies in the image of $k[Y] \to k[X]$ precisely when $f$ lies in $\phi^*(\mathfrak{m}_{\phi(u)})$, which in turn implies that $f$ must vanish on the whole fiber over $\phi(u)$, so that fiber lies in $Z$. That is why we rule out this possibility by assuming that $Z$ does not contain the fiber over $u$.
It is fairly straightforward to now deal with the more general case of a $k$-linear combination of positive powers of $f$ lying in $k[Y] \to k[X]$ as well.

If you get a chance, could you add more details for the sentence: "Now, I think one can check that $f$ lies in the image of $k[Y] \to k[X]$ precisely when $f$ lies in $\phi^*(\mathfrak{m}_{\phi(u)})$"?

The argument I had in mind for that part is as follows. If $f = \phi^*(g)$, then$$f(u) = \phi^*(g)(u) = g(\phi(u)),$$but recall that $f$ lies in $\mathfrak{m}_u$ so $f(u) = 0$, thus $g(\phi(u)) = 0$, so $g$ lies in $\mathfrak{m}_{\phi(u)}$, so $f = \phi^*g$ lies in $\phi^*(\mathfrak{m}_{\phi(u)})$.
If $f$ lies in $\phi^*(\mathfrak{m}_{\phi(u)})$, say $f = \phi^*g$ where $g(\phi(u)) = 0$, then given any point $p$ in the fiber over $\phi(u)$, we get$$f(p) = \phi^*(g)(p) = g(\phi(p)) = g(u) = 0,$$where we used that $\phi(p) = \phi(u)$ by the assumption on $p$. Thus $f$ vanishes on the fiber over $\phi(u)$.
