Pullback of the normal cone The introduction to Fulton's Intersection Theory reads in part:

To give an idea of the main thrust of the text, we sketch what we call the basic construction. [...] To a closed regular embedding $i : X \to Y$ of codimension $d$, and a morphism $f : V \to Y$, with $V$ a $k$-dimensional variety. [...] this construction produces a rational equivalence class of $(k-d)$-cycles on $W = f^{-1}(X)$.  This intersection class $X \cdot_Y V$ can be defined as follows: Since $i$ is a regular embedding, the normal cone to $X$ in $Y$ is a vector bundle; let $N$ denote the pull-back of this bundle to $W$.  The normal cone $C$ to $W$ in $V$ is a $k$-dimensional closed subscheme of $N$. [...]

I'm confused as to why we can regard $C$ as being contained in $N$.  In particular, I can see why we should expect there to be a map from $C$ to $N$ -- heuristically, if we move a little ways in $V$ in a direction normal to $W$ then we can send this over to $Y$ and we shouldn't be moving in a direction parallel to $X$, as $W$ is the preimage of $X$ -- but I can't see why this map ought to be injective.
 A: First of all, let us fix the notation. Let $I\subset \mathcal O_Y$ be the ideal of $X$ in $Y$ and $J\subset \mathcal O_V$ the ideal of $W$ in $V$. For a regular embedding $X\to Y$, the normal cone coincides with the normal sheaf (which is a bundle over $X$, of rank $d$). In other words, the natural surjection $\textrm{Sym }I/I^2\to \oplus_{n\geq 0}I^n/I^{n+1}$ is an isomorphism. Thus $$C_{X/Y}=\textbf{Spec }\bigl(\oplus_{n\geq 0}I^n/I^{n+1}\bigr)\,\,\textrm{and}\,\,N_{X/Y}=\textbf{Spec }\textrm{Sym }I/I^2\,\,\textrm{coincide.}$$
We have the rank $d$ vector bundle $N$ on $W$:
$$N=f^\ast N_{X/Y}\cong \textbf{Spec }f^\ast\bigl(\oplus_{n\geq 0}I^n/I^{n+1}\bigr)
\cong \textbf{Spec } \textrm{Sym }f^\ast(I/I^{2}).$$
The claim is then that $$C=C_{W/V}=\textbf{Spec }\bigl(\oplus_{n\geq 0}J^n/J^{n+1}\bigr)$$
has a closed immersion into $N$. This happens if and only if there is a surjection $$\textrm{Sym }f^\ast(I/I^{2})\twoheadrightarrow \oplus_{n\geq 0}J^n/J^{n+1}.$$
I will not do this in detail, but just set up the local calculation: Let us assume that
$$X=\textrm{Spec }B\hookrightarrow Y=\textrm{Spec }R\,\,\textrm{and}\,\, V=\textrm{Spec }A.$$ 
According to our previous notation, we have $$B=R/I \,\,\textrm{and}\,\, W=\textrm{Spec }A/J,\,\,\textrm{with}\,\,A/J=A/IA=A\otimes_RB.$$ Now, $X\to Y$ being a regular embedding of codimension $d$ implies that $I/I^2$ is a free $B$-module of rank $d$ (generated by the images $x_1,\dots,x_d\in I/I^2$ of a regular sequence $\{a_1,\dots,a_d\}\subset I$ of length $d$. Hence, if we identify $$\textrm{Sym }f^\ast(I/I^2)=A/J[f^\ast x_1,\dots,f^\ast x_d],$$ it should be clear that it surjects onto $\oplus_{n\geq 0}J^n/J^{n+1}$. (Maybe it is useful to write $J^n/J^{n+1}$ as $J^n\otimes_A(A/J)$.)

In general. The result you are after is that if $X\to Y$ is a regular embedding and $f:Z\to Y$ is any morphism, then $f^\ast \mathcal C_{X/Y}$ has a natural sheaf surjection onto $\mathcal C_{f^{-1}(X)/Z}$. (Here $\mathcal C$ denotes the conormal sheaf.) Moreover, this is an isomorphism when $f$ is flat.
