Splitting is not natural in UCT

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?

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.)