The dimension of the component of a variety Mumford claimed the following result:
If $X$ is an $r$-dimensional variety and $f_{1},...,f_{k}$ are polynomial functions on $X$. Then every component of $X\cap V(f_{1},...,f_{k})$ has dimension $\ge r-k$. 
He suggested this is a simple corollary of the following result:
Assume $\phi:X^{r}\rightarrow Y^{s}$ is a dominating regular map of affine varieties. Then for all $y\in Y$, the dimension of components of $\phi^{-1}(y)$ is at least $r-s$. 
However it is not clear to me how the two statements are connected. Let $Y=X- V(f_{1},...,f_{k})$ where the dominating regular map is the evaluating map by $f_{i}$. Then $Y$ should have dimension at least $r-k$ by the definition. So in particular the components of $o$'s preimage should have dimension at most $r-(r-k)=k$, and least $r-r=0$. I feel something is wrong in my reasoning and so hopefully someone can correct me. 
I realized something may be wrong in my reasoning: Notice two extremes $\dim(Y)$ can be by letting $f_{i}$ be one of the generating polynomials of $\mathscr{U},X=V(\mathscr{U})$, in this case $\dim(Y)=0$. And on the other hand if $f_{i}$ does not vanish at all on $X$, then $\dim(Y)=r$. So $Y$ seems not a good choice as it is insensitive to the value to $k$. 
However, a reverse way of reasoning might be possible by using $Y=X\cap V(f_{1},..f_{i},f_{k})$ while claiming the component can only have dimension at most $k$. Then by inequality we have the desired result. But I do not see how $X\rightarrow Y$ by quotient map to be a regular map. 

To deduce the corollary, let $Y$ be $k$-dimensional affine space, and let $\phi$ be the map sending a point $x \in X$ to $(f_1(x), f_2(x), \ldots, f_k(x))$.  Then $X \cap V(f_1, \ldots, f_k)$ is the preimage $\phi^{-1}(0,0,\ldots,0)$, hence, by the result, its dimension is at least $r-k$.
(The fact that $\phi$ might not be dominating is unimportant: We can always replace $Y$ with the image of $\phi$.)
