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Suppose we have four spectra: $E_1$, $E_2$, $X_1$ and $X_2$ where $X_i$ is $E_i$ local and $L_{E_1\wedge E_2}X_1\simeq L_{E_1\wedge E_2}X_2$. Of course what this really means (I think), if we're thinking in terms of model categories, is that there exists an $E_1\wedge E_2$-equivalence between $X_1$ and $X_2$. If we construct (perhaps after fibrant approximation?) the homotopy pullback of the diagram $X_1\to L_{E_1\wedge E_2}X_1\leftarrow X_2$, denoted, say, by $X^{1,2}$. Then is it true that there is an $E_1$-equivalence between $X^{1,2}$ and $X_1$ as well as an $E_2$-equivalence between $X^{1,2}$ and $X_2$, or in other words, $X^{1,2}$ localizes to $X_1$ and $X_2$ along the obvious localization functors? Feel free to make as many assumptions as you'd like, i.e. both localizations are smashing, everything is fibrant-cofibrant, some particular model of spectra, whatever.

The intuition here is that the $E_1$-local stuff of $X_2$ should somehow be the same as the $E_2$-local stuff of $X_1$ and we'd like to glue along that part in some way.

Forgive me if I'm messing up the details or the fundamentals (perhaps certain things should be up to homotopy or not up to homotopy, or certain things should be fibrantly or cofibrantly replaced or something). If it is clear to anyone what I am trying to say, but that person can say it in the correct formal way, please help!

Thanks!

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Nah. I'm thinking of it as "intersection" in the locale of Bousfield classes DL. –  Jon Beardsley Apr 20 '12 at 0:28
    
Although I think in special cases that pullback should be just the wedge product. I'm not sure how it would all work either, in term of what the pushout of the wedge along localizations would look like. –  Jon Beardsley Apr 20 '12 at 0:44
    
@Jon: Perfectly right, that was my mistake. –  Tyler Lawson Apr 20 '12 at 0:48
    
I suspect it's something like the fact that all $E_1$-acylics that map to $X_1$ also map to something which is equivalent to $X^{1,2}$, and so at least their acyclizations are equivalent. Still working on it though. –  Jon Beardsley Apr 20 '12 at 16:16

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