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Let us note $\Delta_p(X)$ the $p$-singular chains on a topological space $X$. We have a well-known barycentric subdivision $$b:Δ_p(X)→Δ_p(X).$$ Is $b$ injective ? Moreover, does $b$ have a retraction ? I think I can prove the injectivity if $X$ is a manifold, using integration of differential forms, but in the general case it is not very clear.

If necessary, one can assume that $X$ is Hausdorff and localy compact.

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No: consider the circle. Let $\sigma_1$ be the singular 1-simplex that wraps exactly once counterclockwise around the circle, and let $\sigma_2$ be obtained from $\sigma_1$ by reversing the orientation and rotating 180 degrees. Then $\sigma_1+\sigma_2$ is nontrivial, but the barycentric subdivision consists of four 1-simplices that cancel in pairs.

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  • $\begingroup$ If I'm not mistaken, the cancellation in pairs occurs in homology but not in $\Delta_p$. $\endgroup$
    – C. Dubussy
    Commented Mar 27, 2017 at 11:59
  • $\begingroup$ But if we don't reverse the orientation and only do a rotation of 180 degrees, then $\sigma_1 \neq \sigma_2$ and $b(\sigma_1)=b(\sigma_2).$ $\endgroup$
    – C. Dubussy
    Commented Mar 27, 2017 at 13:14
  • $\begingroup$ If you don't reverse the orientation, nothing can cancel. $\endgroup$
    – Greg Friedman
    Commented Mar 28, 2017 at 19:27
  • $\begingroup$ And I'm still pretty sure the cancellation happens as I originally said. Look, suppose $\sigma_1$ has its endpoints at $(1,0)$, embedding the circle in the plane in the standard way. Then the barycentric subdivision of $\sigma_1$ is the sum of the two CCW-oriented arcs between $(1,0)$ and $(-1,0)$. Similarly, $\sigma_2$ has its endpoints at $(-1,0)$, and so is not $-\sigma_1$, but its subdivision is the sum of the two arcs oriented clockwise. So everything cancels as singular chains. $\endgroup$
    – Greg Friedman
    Commented Mar 28, 2017 at 19:29

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