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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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$G$ is supposed to be already defined on the image of $R$, and then it is extended using the formula $G(x)=(H(x),G(R(x)))$ (here $G(R(x))$ is already known) –  user8268 Apr 22 '12 at 16:36
    
It makes sense alright. The author assumes $G$ to be already defined on the outside walls of the cube $I^n$ but not on the upper lid. To define the map on the entire cube, he starts by collapsing the cube onto the hollow box where $G$ is already defined,and then uses the already existng part of $G$ to send the cube into the total space of the fibering. –  Olivier Bégassat Apr 22 '12 at 16:39
    
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. –  Dan Ramras Apr 23 '12 at 5:17

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