Let $B_n$ be the braid group of $n$ strings over the unit disk $D$. Let $$d_i:B_n\to B_{n-1}$$ be the operation which is obtained by forgetting the $i$-th strand, $1\leq i\leq n$. Geometrically this "forgetting operation" is clear.
However, if we use the Artin presentation of the braid group; namely, define $B_n$ as the group with generators $\sigma_1,\cdots,\sigma_{n-1}$ satisfying the following relations:
Then how to describe the "forgetting operation" $d_i$?
Edit: This is totally not correct, although I will leave it up for posterity's sake.
It is clear that this can't be right; as suggested in the comments, if we look at $B_3$ which has two generators $\sigma_i$ satisfying $$ \sigma_1\sigma_2\sigma_1 = \sigma_2\sigma_1\sigma_2 $$ then if we "forgot" the last strand as I suggested by setting $\sigma_2 = 1$, then this would yield a group with one generator $\sigma_1$ satisfying $\sigma_1^2 = \sigma_1$, i.e. the trivial group. So this is clearly not correct.
Original answer
If you think of the generators $\sigma_i$ as swapping the $i$-th string and the $(i+1)$-st string, then it should be easy to see that forgetting the $n$-th strand is equivalent to setting $\sigma_{n-1} = 1$ (since if you forget that string, twirling the previous string around... nothing... does nothing).
So at least in the simplest case (strings at either end, really), this is what the forgetting map does. I believe (though I haven't bothered with the details) that forgetting other strands is similar, with some sort of conjugacy argument.