# Is Transversality invariant by losing Dimension?

Setup Let $X$ be a smooth, reduced and irreducible scheme. And let $E$ be a vector bundle of rank $n$ on $X$. Following Eisenbud and Harris we say that the collection $\{\sigma_1,...,\sigma_k\}$ of sections of $E$ is dimensionally transverse if $codim_X(V(\sigma_1\wedge ...\wedge \sigma_k))=n-k+1$.

Here $V(\sigma_1\wedge ...\wedge \sigma_k):=\{x\in X:\sigma_1(x)\wedge...\wedge\sigma_k(x)=0\}$, i.e. it is the degeneracy locus of the collection of sections. By a theorem of Macaulay we know that the codimension above is, in any case, at most $n-k+1$. Moreover if the codimension is $n-k+1$, then it is pure, i.e. any connected component has the same codimension.

The construction above is motivated by the fact that if we have such a collection of sections, then we can define the $k$-th Chern Class of $E$ by $c_k(E)=[V(\sigma_1\wedge...\wedge\sigma_k)]$.

Eisenbud and Harris claim that if we have a collection of $n$ dimensionally transverse sections of $E$ then we can define the Total Chern Class of $E$: $c(E)=1+c_1(E)+...+c_n(E)$. In specific it means that if we have a collection of $n$ dimensionally transverse sections, then we can find a collection of $k$ dimensionally transverse sections for any $k\leq n$.

Unfortunately I don't see how to prove (or disprove) it, may you please help?

Advancement I tried to prove that cutting out a section from a collection of dimensionally transverse sections would give a collection of dimensionally transverse sections. But, though I don't have a counterexample, it does not seem to be the case.

Moreover it is easy to observe that, if there is a counterexample to the sentence in bold, then the vector bundle must not be globally generated.

Edit This is the restyling of a question which didn't appear, in my opinion, to receive enough attention. If you have any constructive criticism or suggestion on how to improve it I would be very glad to receive them!

Thank you for your time!

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