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Sound: Grey Francolin Song Grey Francolin Wav Download: Listen to Grey Francolin Bird Calls: Grey Francolin. For more Grey Francolin Bird Calls, Grey Francolin Bird Sounds, Grey Francolin Birds, Grey Francolin Noises sample mp3 ringtone.Q: A question about cohomology of homotopy colimit Sorry if it's a basic question, but I can't find my answer in Hatcher's book. Let $X_{\bullet}$ be asimplicial set and $X$ be the homotopy colimit of $X_{\bullet}$ (in the projective model structure on simplicial sets). Denote $p^*(X)$ as the pullback of $X$ to a point. From the definition of homotopy colimit, there is a canonical map $p_n: X_n \to p^*(X)$ for each $n$. Now, when we look at the sheaf cohomology, we get $H^i(p^*(X);R) = H^i(X;R)$ for each $R$, and $H^i(X;\mathbb{Z})\simeq H^i(p^*(X);\mathbb{Z})$ for each $i$. Why is it true that $H^i(X;R) = H^i(p^*(X);R)$ for each $R$? Edit: $\Pi_n(X)$ is the free profinite group on $n$ points. A: The answer is because $H^n(X;R)$ is a projective module (forget about sheaves). To compute projective modules, it is enough to compute projective modules over the rings of the form $\displaystyle R \oplus \bigoplus_{i \in I}\mathbb Z/p^k$. Now, the functor $p^*(-)$ takes the form of $\mathbb Z/p^k \oplus \bigoplus_i \mathbb Z/p^k$ and thus preserves projective modules. ,n} & = \frac{1}{n^2} \Bigg[ \sum_{i=1}^n \sum f678ea9f9e
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Grey Francolin Bird Sound Mp3