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In this post, it says that 'the universal coefficient theorem does not a priori imply that the induced map is trivial because the splitting $H^i(X) = \operatorname{Hom}(H_i(X),\Bbb Z)\oplus \operatorname{Ext}(H_{i-1}(X),\Bbb Z)$ is not natural. Does that mean that the following diagram not necessary commute? . so that $f^*$ is not necessary trivial?

one potato two potato
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    That sounds about right – Jeroen van der Meer Dec 02 '21 at 15:05
  • See also https://math.stackexchange.com/q/33685/172988 (naturality would be that you can choose those horizontal arrows in a consistent way so that for all choices of $X$ and $Y$ the diagram commutes). – Kyle Miller Dec 02 '21 at 18:04
  • @KyleMiller So if the splitting is natural, there is some horizontal map (iso) so that the diagram commutes but in my case, splitting is not natural so there is no horizontal map that makes the above diagram commute. Right? – one potato two potato Dec 03 '21 at 02:48
  • @love_sodam Almost: "The splitting is natural" means that the horizontal arrows are a function of $X$ and $Y$. You might still accidentally get a commutative diagram (although it's not always possible -- Hatcher gives an example as an exercise). – Kyle Miller Dec 03 '21 at 04:18

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When comparing two things which are isomorphic, you should not write that they are equal. This will inevitably lead to confusion. To compare them, you need to specify an isomorphism between them.

Indeed, the horizontal arrows are not well-defined. The isomorphism you are talking about depends on many choices. It doesn't even really make sense to say this diagram commutes without first specifying the horizontal arrows.

The stronger claim is that there is no way to define the horizontal arrows for all $X$ so that this diagram always commutes. (This is also true, but hardly obvious.)

user1000169
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