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Let $\mathcal T$ be a triangulated category admitting countable coproducts and hence homotopy colimits of sequences $X^0 \to X^1\to X^2\to \cdots$ in $\mathcal T$. Given such a sequence, I would like to know if there exist choices of maps $\text{C}(X^0\to X^n)\to\text{C}(X^0\to X^{n+1})$ such that there exists an isomorphism $\text{C}\left(X^0\to\text{hocolim}(X^n)\right)\cong\text{hocolim}\left(\text{C}(X^0\to X^n)\right)$.

More generally, one could ask the following: Given a map $f^{\ast}: X^{\ast}\to Y^{\ast}$ between two sequences as above, is it possible to choose maps between the cones $\text{C}(X^n\to Y^n)\to\text{C}(X^{n+1}\to Y^{n+1})$ and a map $\text{hocolim}(X)\to\text{hocolim}(Y)$ (yielding a morphism of the defining triangles of the homotopy colimit) such that there exists an isomorphism $\text{C}\left(\text{hocolim}(X)\to\text{hocolim}(Y)\right)\cong \text{hocolim}\left(\text{C}(X^n\to Y^n)\right)$?

Any ideas?

Thank you very much!

share|improve this question
    
cofibers are homotopy colimits. This is just a moral argument. –  Sean Tilson Jan 8 '12 at 22:28
    
Hi Sean! Ok, so one might even ask if homotopy colimits commute with homotopy colimits a sense as above. There is such a question on MO, but it seemed that there only triangulated categories with models are considered and I couldn't follow. Do you have an idea on how to rigorously prove maybe at least the first statement? –  Hanno Becker Jan 9 '12 at 6:44

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