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I'm trying to understand the last bit of the proof. But, it makes no sense as if we defined $G=(H(x), G(R(X))$ won't we get some crazy recursive function. Does he mean $G=(H(x),R(x))$. Also, is there a way to make the induction more rigorous. It just seems to be skimpy and then the last bit shocked me as I don't see how it's valid.

As you would get something crazy like this $G=(H(x),(H(x),(H(x), \ldots )\ldots))$

The paper is this one http://www.math.uchicago.edu/~amwright/HomotopyGroupsOfSoheres.pdf

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    This discussion is pretty clearly about the homotopy *lifting* property and not the homotopy extension property, so I guess there's a typo in the statement of the Proposition... Of course this is an application of the homotopy extension property for cubes relative to certain faces.2012-04-23

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