In Allen Hatcher's book Algebraic topology he states that factoring out 2-torsion $\pi_{i}(S^{2n})\cong\pi_{i-1}(S^{2n-1})\times\pi_{i}(S^{4n-1})\:\forall n$ but in his book Spectral Sequences in Algebraic Topology he has $\pi_{i}(S^{n})\cong\pi_{i-1}(S^{n-1})\bigoplus\pi_{i}(S^{2n-1})\:\forall n$ even. Am I correct in assuming these are the same or am I totally missing something?
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2 Answers
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Finite direct sums and finite direct products are the same thing on groups. The homotopy theory part has nothing to do with it. (Infinite direct sums and products are different, though, so be careful about this.)
answered Mar 2, 2014 at 18:46
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$\begingroup$ The difference is this: An infinite sum of groups has elements with only finitely many coordinates non-zero. An infinite product does not have this limitation on its elements (the real, formal difference is category-theoretical, but for groups, this is what the difference manifests as) $\endgroup$– ArthurMar 2, 2014 at 18:47
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The higher homotopy groups are all abelian, so for finite indices you have the canonical isomorphism:
$$ \prod^{n}_{i=1}A_{i}=\bigoplus^{n}_{i=1}A_{i} $$ by mapping $(a_{1}\cdots a_{n})$ to $\sum_{i=1}^{n}a_{i}$.
answered Mar 2, 2014 at 18:46
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$\begingroup$ Abelianness is unnecessary here. (The direct sum still makes sense for nonabelian groups, it just isn't the coproduct anymore. Instead it's the "commutative coproduct.") $\endgroup$ Mar 2, 2014 at 19:06
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$\begingroup$ Yes, thanks for the reminder. I am only aware of direct sum as the coproduct. But I think you mean "non-commutative coproduct" right? $\endgroup$ Mar 2, 2014 at 19:09
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$\begingroup$ No, I mean "commutative coproduct." The direct sum of a family of groups $G_i$ is the universal group equipped with a map from all of the $G_i$ all of whose images commute. $\endgroup$ Mar 2, 2014 at 19:41
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$\begingroup$ I see. Thanks for explaining. This does appear to be the right generalization. $\endgroup$ Mar 2, 2014 at 19:55