Timeline for Exterior product in relative cohomology
Current License: CC BY-SA 3.0
20 events
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| Dec 5, 2015 at 8:00 | comment | added | Gustavo Granja | No problem. If you just want homotopy equivalence, you can also reduce to the initial situation of embeddings by replacing $X$ with the mapping cylinder of $\rho$ and $Y$ with the mapping cylinder of $\eta$. Then $X\times Y$ will be replaced by a homotopy equivalent space but this will not matter as the mapping cone is homotopy invariant. | |
| Dec 5, 2015 at 0:55 | comment | added | Fabio | Thanks a lot! Actually the homotopy push-out is well-behaved with respect to homotopy equivalence, hence it seems to be more suitable to define the product in cohomology. When $\rho$ and $\eta$ are embeddings, the homotopy push-out of $X \times B \leftarrow A \times B \rightarrow A \times Y$ has the same homotopy type of the ordinary push-out (you can shrink the cylinder), hence it should be the correct generalization. I will check if everything works for the problem I am trying to solve. | |
| Dec 5, 2015 at 0:01 | comment | added | Gustavo Granja | I think that if you redefine $\rho \wedge \eta$ to be the natural map from the homotopy pushout of $X \times B \leftarrow A\times B \rightarrow A \times Y$ to $X \times Y$, then $C(\rho\wedge \eta)$ and $C(\rho) \wedge C(\eta)$ are actually homeomorphic. Write both spaces as quotients of $X \times Y \coprod A \times [0,1] \times Y \coprod X \times B \times [0,1] \coprod A \times B \times [0,1] \times [0,1]$. | |
| Dec 4, 2015 at 23:19 | comment | added | Gustavo Granja | Thanks for the clarification. If you take $\rho=\eta$ to be the constant map from $S^1$ to the one point space, then $C(\rho)\wedge C(\eta)$ is $S^4$. If I understand your definition of $C(\rho\wedge \eta)$ then in this example, that space is an interval (the mapping cone of the map from a one point space to a one point space). However, if you switch your definition of $\rho \wedge \eta$ to have domain the homotopy pushout (or double mapping cylinder), then you do get that $C(\rho \wedge \eta)$ is homotopy equivalent to $S^4$. | |
| Dec 4, 2015 at 22:20 | comment | added | Fabio | The relative cone is the mapping cone of the inclusion. In every cone the marked point is the vertex, hence there is a natural marked point in order to define the smashed product. | |
| Dec 4, 2015 at 22:10 | comment | added | Gustavo Granja | What do you mean by the "relative cone" $C(X,A)$? Is it the mapping cone of the inclusion $X\cup CA$? Also the $\wedge$ notation is confusing. Do you mean the smash product when you say $C(\rho)\wedge C(\eta)$? If so, how are these spaces pointed? | |
| Dec 4, 2015 at 20:01 | comment | added | Fabio | Ok, I corrected the text of the question, thank you. Actually I have no reason to think that they have the same homotopy type, I just hope it. If not, I would like to find any way to pass from the cohomology of $C(\rho) \wedge C(\eta)$ to the one of $C(\rho \wedge \eta)$. Now I am trying to see if prop. 9.8 pag. 224 of the book of Strom, "Modern classical homotopy theory", can help. | |
| Dec 4, 2015 at 19:57 | history | edited | Fabio | CC BY-SA 3.0 |
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| Dec 4, 2015 at 19:27 | comment | added | John Klein | @Fabio If what you are saying is correct then you are using incorrect notation: it is more correct to write $A \times Y \cup_{A \times B} X\times B$, even though $A\times B$ isn't a subset of either $A\times Y$ or $X\times B$. I doubt that $C(\rho \wedge \eta)$ has the correct homotopy type unless you assume that $A \to X$ and $B\to Y$ are cofibrations. | |
| Dec 4, 2015 at 15:28 | comment | added | Fabio | I think the domain of $\rho \wedge \eta$ is the push-out of $A \times Y \leftarrow A \times B \rightarrow X \times B$. We consider the disjoint union $(A \times Y) \sqcup (X \times B)$ and identify $(\rho(a), b) \sim (a, \eta(b))$. The map $\rho \wedge \eta$, with codomain $X \times Y$, is $(a, y) \mapsto (\rho(a), y)$ and $(x, b) \mapsto (x, \eta(b))$. I would like to show that $C(\rho \wedge \eta)$ has the same homotopy type of $C(\rho) \wedge C(\eta)$. | |
| Dec 4, 2015 at 2:35 | comment | added | John Klein | @Fabio : I was confused by your notation, which doesn't seem to be standard. I guess the standard way to write it is this: let $P$ be the pullback of the maps $X\times B \to X \times Y$ and $A\times Y \to X\times Y$. Then your space is the pushout of the diagram $X\times B \leftarrow P \to A\times Y$. Right? | |
| Dec 3, 2015 at 4:01 | comment | added | Fabio | The domain is $(X \times B) \sqcup_{X \times Y} (A \times Y)$, I corrected the text of the question. This means that $(x, b) \sim (a, y)$ if and only if $(x, \eta(b)) = (\rho(a), y)$. When $\rho$ and $\eta$ are embeddings, it coincides with $(X \times B) \cup (A \times Y)$. | |
| Dec 3, 2015 at 4:00 | history | edited | Fabio | CC BY-SA 3.0 |
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| Dec 3, 2015 at 0:42 | comment | added | John Klein | Tom: there are no maps $X\times Y \to X \times B$, $X\times Y \to A\times Y$. I think the $X\times Y$ should be $A\times B$. | |
| Dec 3, 2015 at 0:40 | history | edited | John Klein | CC BY-SA 3.0 |
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| Dec 3, 2015 at 0:34 | comment | added | Tom Goodwillie | What is the domain of $\rho\wedge\eta$? Is it simply the pushout of $X\times B\leftarrow X\times Y\rightarrow A\times Y$? If so, then it is not what you want in general. | |
| Dec 3, 2015 at 0:30 | history | edited | Tom Goodwillie | CC BY-SA 3.0 |
fixed typo (swapped $B$ and $Y$)
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| Dec 2, 2015 at 16:20 | history | edited | Fabio |
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| Dec 2, 2015 at 13:56 | history | edited | Fabio | CC BY-SA 3.0 |
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| Dec 2, 2015 at 13:45 | history | asked | Fabio | CC BY-SA 3.0 |