# when cup product is a zero homomorphism [closed]

How to see that the cup products vanish on suspensions?

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## closed as off-topic by Ricardo Andrade, Andrey Rekalo, Olivier Benoist, Stefan Kohl, Carlo BeenakkerDec 21 '13 at 14:16

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question does not appear to be about research level mathematics within the scope defined in the help center." – Ricardo Andrade, Andrey Rekalo, Olivier Benoist, Stefan Kohl, Carlo Beenakker
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When asking questions like this one, you would do well to add some background and motivation: why do you want to know? Where have you already looked to find an answer (especially given that this sort of thing is covered in many introductory books)? What's wrong with the explanations in those books? – Loop Space Aug 4 '10 at 15:32
I agree with everything Andrew has said, and am voting to close since the question doesn't (in its present form) give any indication of how much work the questioner has put into the question, nor at what level a helpful answer needs to be pitched. – Yemon Choi Aug 4 '10 at 17:58

13.66 in Switzer's Algebraic Topology: Homotopy and Homology. The idea is to use the fact that $\Sigma X$ decomposes into two copies of $CX$, say $A$ and $B$, glued along the common boundary of $X$. For any two cohomology classes $x$ and $y$ in $\tilde{E}^* \Sigma X$, you can uniquely pull $x$ back to a class $x'$ on the relative pair $(\Sigma X, A)$ and $y$ back to a class $y'$ on $(\Sigma X, B)$. Cupping is natural w.r.t the two relative inclusions $i_A: (\Sigma X, \{x_0\}) \to (\Sigma X, A)$ and $i_B: (\Sigma X, \{x_0\}) \to (\Sigma X, B)$, and so you get the calculation $x \smile y = i_A^*(x') \smile i_B^*(y') = i^*(x' \smile y')$, where $i: (\Sigma X, \{x_0\}) \to (\Sigma X, \Sigma X)$ is another relative inclusion and $x' \smile y'$ a class on the pair $(\Sigma X, \Sigma X)$ --- but that guy has trivial reduced cohomology.

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Is it possible that some of your $X$es should be $\Sigma X$es? – darij grinberg Aug 4 '10 at 15:53
Yes, it certainly is :) – Eric Peterson Aug 4 '10 at 15:58

The cup product is a gussied-up version of the map induced by the reduced diagonal map $\bar \Delta$, which is the composite of the ordinary diagonal $\Delta: X\to X\times X$ and the quotient map $q: X\times X \to X\wedge X$; note that $q$ is the cofiber of the inclusion $i: X\vee X\to X\times X$.

If $X$ is a suspension, then the map $\Delta$ has a lift (up to homotopy) $\lambda: X\to X\vee X$ through the inclusion $i : X\vee X \to X\times X$ of the wedge, and hence $\bar \Delta \simeq q\circ \Delta \simeq q\circ i \circ \lambda \simeq *$.

One way to see the lifting is to lift the adjoint; and this is easy because $\Omega(X\vee X) \to \Omega (X\times X)\cong (\Omega X) \times (\Omega X)$
has a homotopy section given by $s: (\omega, \tau) \mapsto (i_1\circ \omega) * (i_2\circ \tau)$, where $i_1, i_2: X\to X\vee X$ are the inclusions of the summands and $*$ denotes concatenation of paths.

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This is a special case of the fact that the cup-length is a lower bound for the Lusternik–Schnirelmann category. Using those two terms as keywords should get you the standard arguments.

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Neat, didn't know this had a name. – Eric Peterson Aug 4 '10 at 15:52