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Given a fibered surface $X \rightarrow C$, with generic fiber $Y$ and a vector bundle $E$ on $X$.

Then the first Chern class $c_1(E)$ is a divisor on $X$, so one can restrict this divisor to the generic fiber $c_1(E)_{|Y}$.

On the other hand one can restrict $E$ to the generic fiber and look at its first Chern class as a divisor on $Y$, i.e. $c_1(E_{|Y})$.

Do we always have the equality $c_1(E)_{|Y}=c_1(E_{|Y})$ as divisors on $Y$, if $X$, $C$ and $Y$ are "nice" enough?

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Yes, even when $X$, $C$ and $Y$ are as nasty as they can be. Chern classes are functorial.

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  • $\begingroup$ Ah thanks. So let me see if i got this: I take the inclusion $i : Y \rightarrow X$, then one has $c_1(i^{*}E)=i^{*}c_1(E)$ by functoriality of $c_1$, which is exactly the equality i'm looking for. $\endgroup$
    – TonyS
    Commented May 8, 2010 at 17:00
  • $\begingroup$ That's exactly right. $\endgroup$
    – Angelo
    Commented May 8, 2010 at 17:08

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