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where $\widetilde{H}$ denotes reduced homology, $\Sigma^k$ denotes suspension and $\Omega^k$ denotes looping.

The context that I need this is to show that the homology groups of a prespectrum $T$ are the same as the homoloy groups of its left adjoint $LT$ as a spectrum (ref. Peter May's A Concise Course in Algebraic Topology pp.233). I propose to prove in the following way

$$H_n(LT) = \text{colim}_{l\to\infty}{\widetilde{H}_{n+l}((LT)_l)} = \text{colim}_{l\to\infty}(\widetilde{H}_{n+l}(\text{colim}_{k\to\infty}{\Omega^k{T_{l+k}}})) $$ $$ \quad = \text{colim}_{l\to\infty}(\text{colim}_{k\to\infty}{\widetilde{H}_{n+l}{\Omega^k{T_{l+k}}}}) = \text{colim}_{l\to\infty}(\text{colim}_{k\to\infty}{\widetilde{H}_{n+l+k}{T_{l+k}}}) = H_n(T)$$

PS: I think I also need it to define the map for reduced homology for a prespectrum like (10.12) in Dan Freed's notes.

PhysicsMath
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  • Did you consider testing your isomorphism for simple spaces? – Pedro Jul 31 '16 at 01:04
  • @Pedro: Do you have simple example or counterexample to share? – PhysicsMath Jul 31 '16 at 01:09
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    Your claim is true for homotopy groups, but the homology groups of loopspaces depend in a more nontrivial way on the homology groups of the original space . Here there are some examples: $H_1(\Omega^2 S^2) $ is nontrivial, but $H_3(S^2)$ is trivial by dimension considerations. – Pedro Jul 31 '16 at 01:33
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    I have a simple counterexample, first $H_n(\sum^k X) = H_{n+k}(X)$ for all $X$, next compute $H_n(\Omega S^k)$ for any $n$ or $k$ and see how it fails. Loops spaces have simple homotopy by definition, but their homology isn't easy. One can compute the homology of loops spaces of spheres using the usual fibration and the serre spectral sequence, and see that this has to be false. – Benjamin Gadoua Jul 31 '16 at 02:22
  • Great comments Pedro and Benjamin! Thank you very much! But then what do you think how I should approach to show $H_n(LT) = H_n(T)$ if the suspected isomorphism in reduced homology is false in general? – PhysicsMath Jul 31 '16 at 04:39

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