So I encountered a term that I don't quite recognize from lecture. The professor stated that a certain short exact sequence splits naturally, but I don't understand what the naturally condition is in this case. Does it means that the squares with the splitting maps in place commutes or what? Note that exact sequence is the exact sequence from the universal coefficient theorem. I know that there is another stackexchange topic on this, but the answer isn't what I was looking for. So I was wondering if my characterization was correct or not.
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That's what I'd read into it. However, Wikipedia tells us that the split in the universal coefficient theorem is not natural. – Hagen von Eitzen Jan 27 '16 at 18:00
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Yeah I understand that the splitting isn't natural, but I was wondering if my characterization of splitting naturally was correct or not. Namely, that such a sequence splits naturally if the commuting squares have maps replaced with splitting maps also commutes. Is this correct? – Enigma Jan 27 '16 at 18:05
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The universal coefficient theorem provides you, for every space $X$, a short exact sequence of abelian groups. This entire short exact sequence is functorial in $X$, which is to say that it can be thought of as a short exact sequence of functors. A natural splitting is a splitting of this short exact sequence of functors: that is, it's a family of splittings which themselves organize into a functor. This means that various squares involving the splitting maps commute as you say.
Qiaochu Yuan
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