This question might be tautological. It comes from a statement in the proof of the non-emptiness of the degeneracy loci of a vector bundle homomorphism that Prof. Lazarsfeld gives in his book "Positivity in AG II" (Theorem 7.2.1)


Take a homomorphism of vector bundles $v:E\rightarrow F$ with kernel $F=\ker v$ and image $K=Im v$ and consider the short exact sequence $$0\rightarrow N \rightarrow E \rightarrow K \rightarrow 0$$


The surjection $E^{\ast}\to N^{\ast}$ gives an embedding $\mathbb{P}(N^{\ast})\hookrightarrow \mathbb{P}(E^{\ast})$ and we seek to realize $\mathbb{P}(N^{\ast})$ as the zero locus of the section of some vector bundle.


The projectibve bundle $\pi:\mathbb{P}(E^{\ast})\rightarrow Y$ comes with a tautological surjection $$\pi^{\ast}E^{\ast}\rightarrow \mathcal{O}\_{\mathbb{P}(E^{\ast})}(1) \rightarrow 0$$


(given by the identity $E^{\ast}\rightarrow E^{\ast}$) and the composition of its dual $\mathcal{O}\_{\mathbb{P}(E^{\ast})}(-1) \rightarrow \pi^{\ast}E$ with the pullback homomorphism $\pi^{\ast}v:\pi^{\ast}E \rightarrow \pi^{\ast}K$ gives a section 


$$s\in \Gamma(\mathbb{P}(E),\pi^{\ast}K \otimes \mathcal{O}\_{\mathbb{P}(E^{\ast})}(1))$$


Then it is claimed that the zero-locus of this section gives precisely the subvariety $\mathbb{P}(N^{\ast})\hookrightarrow \mathbb{P}(E^{\ast})$.


I am wondering whether this is obvious or not, but this correspondence is not apparent to me.


Thanks in advance for any insight.



Projective bundle given by vanishing of a section
It's tautological. Remember that $\mathcal O\_{\mathbb P(E^\*)}(-1)$, viewed as an actual bundle of lines, is the tautological bundle: It is $E$ with the origin blown up. The map to $\pi^\* E$ is literally the map from $E$ with the origin blown up to $E$. So the map to $\pi^\* K$ is that, composed with the projection down to $K$. The section is trivial at a point if the map is trivial in the corresponding line - that is, if the point of $\mathbb P(E^\*)$ corresponds to a line lying in the part of $E$ that maps to $0$ in $K$ - that is, if the line lies in $N \subset E$.


But the points with lines lying in $N \subset E$ are of course just $\mathbb P(N^\*) \subset \mathbb P(E^\*)$.


