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Let $F,E$ be two locally free sheaves of equal rank on a complex manifold $X$.

Assume that we have an injection $0\to F\to E$ such that the induced map $0\to \det F\to \det E$ is an isomorphism. Is then the original map an isomorphism too? If not always, what if also $\det F=\det E=\mathcal{O}_X$ ?

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    $\begingroup$ I think this is true and that it follows from the fact that determinants commutes with taking fibers : if $f_x:F_x\rightarrow E_x$ was not an isomorphism at $x$ (here I mean the fibers of the vector bundles, not the stalks of the associaded sheaves), then $\det f_x:(\det F)_x\rightarrow (\det E)_x$ will not be an isomorphism. $\endgroup$
    – Roland
    Commented Jan 22, 2016 at 13:18
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    $\begingroup$ I don't think it's wise to think in terms of vector bundles rather than sheaves (for this question). Indeed, an "injection of vector bundles of the same rank" is obviously an isomorphism, but this is not true if "vector bundle" is replaced by "locally free sheaf". $\endgroup$
    – Jake Levinson
    Commented Jan 22, 2016 at 14:46
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    $\begingroup$ Thank you @Jake. You are right and this is what I actually had in mind. I will edit $\endgroup$
    – Heitor Fontana
    Commented Jan 22, 2016 at 14:48
  • $\begingroup$ Glad to be of help (it appears we are writing simultaneously :-) ) $\endgroup$
    – Jake Levinson
    Commented Jan 22, 2016 at 14:50

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The question is local, so pass to the stalks, where both $E,F$ can be assumed free.

Then the statement is that a module endomorphism of $R^n$ is an isomorphism iff its determinant is a unit (rather than merely a nonzerodivisor), which is true.

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