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Let $X$ be a smooth (complex) threefold and $\gamma\in {\rm CH}_1(X)$ a homologically trivial $1$-cycle. Is there a way to construct a (singular) surface $S\subset X$ supporting $\gamma$ such that, denoting $\tau:\widetilde S\rightarrow S$ a desingularization, there is a homologically trivial $\widetilde\gamma\in Pic^0(\widetilde S)$ satisfaying $$i_*\tau_*\widetilde \gamma =\gamma \ \ {\rm in\ CH}_1(X)$$ where $i:S\hookrightarrow X$ is the inclusion?

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  • $\begingroup$ You mean $\gamma$ in $\mathrm{CH}^1(X)$ (codimension $1$)? $\endgroup$
    – Qfwfq
    Mar 12, 2018 at 12:26
  • $\begingroup$ Thank you. No, I do mean "curves". $\endgroup$
    – pi_1
    Mar 12, 2018 at 12:38
  • $\begingroup$ If you take any surface containing the curves in the support of $\gamma$, and take a resolution of it, you get a class in $Pic(\widetilde S)$ that pushes forward to $\gamma$ like you say. Is your question whether we can take this to be homologically trivial? $\endgroup$
    – byu
    Mar 12, 2018 at 14:19
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    $\begingroup$ I suspect that this fails already for Hironaka's example, as described in Appendix B of Hartshorne's "Algebraic geometry". $\endgroup$ Mar 12, 2018 at 15:49
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    $\begingroup$ For a projective example, simply take any cycle which is homologically equivalent to zero but not algebraically equivalent to zero. (On a surface, any 1-cycle which is homologically trivial is algebraically trivial.) $\endgroup$
    – naf
    Mar 13, 2018 at 5:36

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