Fadell--Neuwirth exactness: the kernel is contained in the image

The hard inclusion of exactness at the middle term of the Fadell--Neuwirth sequence: a pure braid on $n+1$ strands which becomes trivial after forgetting the last strand comes from a loop of that last strand alone in the plane punctured at the other $n$ points.

In symbols, for the sequence

with the first map induced by the fibre inclusion $\iota$ and the second by the forgetful map $p$, the claim is $\ker p_* \le \operatorname{im} \iota_*$.

The opposite inclusion is elementary and is already proved: the composite $p \circ \iota$ is the constant map at the base configuration, since forgetting the last coordinate of $(1, 2, \ldots, n, z)$ returns $(1, 2, \ldots, n)$ whatever $z$ is. So the composite induces the trivial homomorphism, and the image of $\iota_*$ lies in the kernel of $p_*$ for purely formal reasons. All the content of exactness is in the direction stated here.

The classical proof uses the homotopy lifting property. Given a loop $\gamma$ of configurations on $n+1$ points whose projection is null-homotopic, lift a null-homotopy of $p \circ \gamma$ through the fibration $F_{0,n+1}E^{2} \to F_{0,n}E^{2}$. The lifted homotopy deforms $\gamma$ into a loop lying entirely in a single fibre, and that fibre is the plane punctured at the $n$ fixed points, so the deformed loop is $\iota$ of a loop there.

This is where it is used that the Fadell--Neuwirth projection is a fibration. That input is not available from covering-space theory, which is what distinguishes this statement from the covering-space facts about the ordered-over-unordered projection.