# Image of a map on cohomology rings

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.

• 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. – Qiaochu Yuan Sep 25 at 2:39
• 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). – Connor Malin Sep 25 at 2:54
• I guess you mean $R=f^*(Z)$? – user43326 Sep 25 at 5:55

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.
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.)