Let $X$ be a topological space, $R$ a ring, $R_X$ the locally constant sheaf defined via $$U \to \Gamma(U, R_X)=\{f:U \to R \vert f \text{ locally constant}\}$$

The $r$-th singular cohomology $C^r(X,R)$ is defined via

$$C^r(X,R) = \prod_{f:\Delta_r \to X}R = Hom(\bigoplus_{f:\Delta_r \to X}\mathbb{Z} \to R)$$

We are going to compare singular and sheaf cohomology.

In order to do this we define the presheaf $\bar{I_r}$

$$U\to C^r(U,R)$$

and get $I_r$ via sheafification of $\bar{I_r}$.

The coboundary maps

$$(\partial \phi)(f) = \sum(-1)^i\phi(..., f\vert _{\partial_i \Delta_r},...)$$

defines the complex

$$0 \to R_X \to I_0 \to I_1 \to ...$$

My question why is this complex exact if and only if $X$ is locally contractable?

  • $\begingroup$ One direction is clear. For the other direction, if the complex is exact then there is a covering $U_i$ of $X$, so that $\Gamma(U_i, C^{\bullet})$ is an exact complex. In particular it implies that $H^k(U_i,\Bbb Z) = 0 $ for all $k \in \Bbb N$ and $i \in I$. This is not too far from being locally contractible. $\endgroup$ – Nicolas Hemelsoet Apr 22 '18 at 19:06
  • $\begingroup$ Hi. Thank you for the answer. One request: You wrote that the direction loc contractable to exact is clear... how to see it? $\endgroup$ – KarlPeter Apr 22 '18 at 19:22
  • $\begingroup$ It's because if $X$ is locally contractible, then there is a covering $U_i$ with $\cup U_i = X$ and each $U_i$ is contractible. And for contractible space, the cohomology complex is exact (i.e for a contractible space $U$ we have $H^k(U) = 0$ for all $k > 0$). $\endgroup$ – Nicolas Hemelsoet Apr 22 '18 at 19:27
  • $\begingroup$ ...ah yes I see it. This is locally the cohomology of $D^0$ because of homotopy invariance and locality of the problem. Thank you. $\endgroup$ – KarlPeter Apr 22 '18 at 19:30
  • $\begingroup$ Yes. I still don't know how to prove the other direction though. $\endgroup$ – Nicolas Hemelsoet Apr 22 '18 at 19:31

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