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16:12
In Calculus II, Goodwillie shows that if an S-cube X is strongly cocartesian, and X(0) --> X(s) is k_s-connected for all s, then the "product" cube X^m is k-cocartesian, where k depends on the k_s and the cardinality of S. The proof reduces to calculating the connectivity of a CW pair. Does anyone know of a more "modern" interpretation that doesn't involve this CW pair argument?
 
2 hours later…
17:49
@NiallTaggart Where is that statement in the paper? And what is m?
@CharlesRezk it’s in the proof of Example 4.4, and m is just some natural number, so the “product” cube sends U to X(U)^m
Isn't this just the usual fact that if $A\to A'$ is a-connected, and $B\to B'$ is b-connected, then $(A\times B')\cup_{A\times B} (A'\times B)$ is $(a+b+1)$-connected? ($\pm \epsilon$)

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