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The following seems like an extremely basic algebraic topology question, but it's not something I ever learned, nor does it look familiar to the algebraic topologists I've asked.

Let $f:X\to Y$ be a map, inducing $f^*:H^*(Y)\to H^*(X)$. Hence the image $R$ of $f^*$ is a subring of $H^*(X)$. Is there a natural way to factor $f$ as $X \to Z \to Y$, such that $f^*$ factors as $H^*(Y) \twoheadrightarrow R \hookrightarrow H^*(X)?$

I have a particular $X,Y$ in mind (the inclusion of one compact complex manifold into another, each with even-degree cohomology) but I'm hoping phrases like "Postnikov tower", "cofibrant replacement", "mapping cone" will serve to give a general answer.

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    $\begingroup$ I wonder if you’d have more luck asking for some kind of image factorization for rational spaces, which presumably could be connected to some kind of image factorization for Sullivan models. $\endgroup$
    – Qiaochu Yuan
    Commented Sep 25, 2019 at 2:39
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    $\begingroup$ I have the suspicion that this is only true for disjoint unions of Eilenberg-MacLane spaces (as the codomain). Or perhaps pairs of spaces where cohomology detects the difference between maps (as is true for any pair with codomain an Eilenberg-MacLane space). $\endgroup$
    – Connor Malin
    Commented Sep 25, 2019 at 2:54
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    $\begingroup$ I guess you mean $R=f^*(Z)$? $\endgroup$
    – user43326
    Commented Sep 25, 2019 at 5:55

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No. Consider the Hopf map $\eta:S^3\to S^2$. If there were such a space $Z$, it would have $\widetilde H^*(Z)=0$, so at the very least $Z$ would be stably trivial, forcing $\eta$ to be stably trivial; but it’s not.

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    $\begingroup$ Well that was straightforward! I'm hoping someone has a more positive answer (i.e. extra conditions to require to make it possible), so I won't accept this quite yet, but it's certainly a definitive answer to the question as stated. $\endgroup$
    – Allen Knutson
    Commented Sep 25, 2019 at 0:05
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One special case of your set up is when $Y=X$ and $f^*$ is idempotent: $f^* \circ f^* = f^*$. In this case, let $Z$ be the mapping telescope of $X \xrightarrow{f} X \xrightarrow{f} X \rightarrow \dots$. This comes with a canonical map $r: X \rightarrow Z$ such that $r^*$ is monic with image equal to the image of $f^*$, and in many cases, one can show that there exists $i: Z \rightarrow X$ such that $r^* \circ i^* = f^*$.

(We are basically looking to lift an idempotent in homology to an idempotent in homotopy. One sufficient condition for $i$ to exist is that $X$ be a $p$--complete CW complex of finite type: see the short paper Atomic spaces and spectra I wrote with J.F. Adams back in the late 1980's.)

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