Suppose $C^{(d)}$ is the $d$-th symmetric product of a curve, embed it in $C^{(d+s)}$ by $E\to E+sP_0$ where $P_0$ is a fixed point. The normal bundle of this embedding is denoted by $N$.
Suppose $|D|\cong \mathbb{P}^r$ is the subscheme of $C^{(d)}$ consisiting of divisors linear equivalent to some divisor $D$.
Is the embedding a regular embedding? How do we show $N|_{|D|}$ has Chern class $(1+h)^s$?($h$ is $O(1)$ on $|D|$).(Fulton's Interseciton theory Example 4.3.2)
Here is an answer for the question: for the case $s=1$, we choose a point $Q$ on $C$ and have embedding $i_Q\colon C^{(d)}\to C^{(d+1)}$ sending $D\to D+Q$, also we have $i'_Q\colon C^{(d)}\to C^{(d)}\times C$, finally the natural map $\pi\colon C^{(d)}\times C\to C^{(d+1)}$ makes $i_Q=\pi\circ i'_Q$.
We consider the following diagram
$\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} 0 & \ra{} & T_{C^{(d)}} & \ra{} & {i'_{Q}}^*T_{C^{(d)}\times C} & \ra{} & N_{C^{(d)}/C^{(d)}\times C}\cong O_{C^{(d)}} & \ra{} & 0 \\ & & \da{} & & \da{} & & \da{} & & & \\ 0 & \ras{} & T_{C^{(d)}} & \ras{} & i_Q^*T_{C^{(d+1)}} & \ras{} & N_{C^{(d)}/C^{(d+1)}} & \ras{} & 0\\ \end{array}$
Thus we have $c_1(N)-c_1(O)=c_1(i_Q^*T_{C^{(d+1)}})-c_1({i'_{Q}}^*T_{C^{(d)}\times C})$. It equals $c_1(i_Q'^*\pi^*T_{C^{(d+1)}})-c_1({i'_{Q}}^*T_{C^{(d)}\times C})=i_Q'^*(c_1(\pi^*T_{C^{(d+1)}})-c_1(T_{C^{(d)}\times C}))$.
But by calculating the ramification of $\pi$ we have $\pi^*K_{C^{(d+1)}}=K_{C^{(d)}\times C}-\Delta$, where $\Delta$ is the reduced divisor having of two same points. Recall that $c_1(T)=-c_1(K)$ (by calculating chern roots). We finally have $c_1(N)=i_Q'^*\Delta$, which restricts to the linear subsystem $\{D|Q\in \textrm{Supp}D\}$, which is a hyperplane. So the chern class is $1$.