Let's fix a smooth integral algebraic variety $X$ over $\mathbb C$. If $\mathscr E$ is a locally free sheaf on $X$, then at each closed point $x\in X$ we have a complex vector space $E_x:=\mathscr E_x\otimes\mathbb C$ and $E:=\bigsqcup_x E_x$ can be endowed with a structure of smooth vector bundle. The transition between locally free sheaves and vector bundles is functorial and it is actually an equivalence of categories.

Now assume that $\varphi:\mathscr E\to\mathscr H$ is an injective morphism between locally free sheaves. It means that for any $x\in X$ we have an injective morphisms of $\mathscr O_{X,x}$-modules $\varphi_x:\mathscr E_x\to\mathscr H_x$.

It is well known that the induced map at the level of vector spaces (fibres of vector bundles) $\Phi_x:E_x\to H_x$ is in general not injective. This because the operation $-\otimes\mathbb C$ doesn't preserve left exactness.

Finally the question: Let's keep the above hypotheses and notation. Is it true or false that there is always a Zariski open subset $U\subset X$ such that $\{\Phi_x\}_{x\in U}$ are injective (as homomorphism of vector spaces)? Are we able to describe such a subset? Is it characterized by some "nice" property?


1 Answer 1


There are many ways of deciding where $\Phi_x$ is injective. Here is one. The injectivity of $\varphi$ implies $r$, the rank of $\mathcal{E}$, is less than or equal to that of $\mathcal{H}$ and $\theta:\wedge^r\mathcal{E}\to \wedge^r\mathcal{H}$ is injective. Easy to check that $\Phi_x$ is injective if and only if the similar map induced by $\theta$ is injective and thus we may reduce to the case when $\mathcal{E}$ is a line bundle. Twisting by its inverse, we may further assume that $\mathcal{E}=\mathcal{O}_X$. Thus, the closed set where the map is not injective is precisely where this section vanishes.


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