# Which singular homology classes can be represented by embedded manifolds?

Given a connected CW-complex $$X$$ I'm interested in if a given homology class $$\sigma \in H_n(X)$$ can be represented by a manifold meaning if there is a map $$f : M^n \to X$$ from a oriented manifold $$M$$ for which $$f_*([M^n]) = \sigma$$. Obviously this is always true for $$n = 1$$ and I could prove it for $$n = 2$$, but it seems this doesn't hold for any $$n$$.

For example I found this answer which talks about the case where $$X$$ is itself a manifold. It says there are cases where $$\sigma$$ is not represented by a manifold for $$n = 7$$. Are there similar results for $$X$$ that are not necessary manifolds?

I'm especially interested in the simpler case where $$H_i(X) = 0$$ for $$1 < i < n$$

The question in the title differs from the question spelled out in the post: In the title, you ask for embedded manifolds, in the post you ask for just maps from manifolds. I think the version of the question asking for embedded manifolds but $$X$$ an arbitrary CW-complex is not very well-behaved, so let me answer the question in the post.
One way to think about this is that there is also a homology theory based on oriented manifolds mapping to $$X$$, called oriented bordism, $$\operatorname{MSO}_*(X)$$. The construction which assigns to a class represented by an oriented manifold with map to $$X$$ the image of its fundamental class in $$H_*(X)$$ comes as a natural transformation $$\operatorname{MSO}_*(X) \to H_*(X)$$ of homology theories. In fact, it lifts to a map of spectra, $$\operatorname{MSO}\to H\mathbb{Z}$$, and this is the bottom map in the Postnikov tower for $$\operatorname{MSO}$$. This way, the question of when homology classes of $$X$$ are in the image of this natural transformation relates this to differentials in the Atiyah-Hirzebruch spectral sequence for $$\operatorname{MSO}_*(X)$$. The existence of homology classes which are not in the image corresponds to the fact that the map $$\operatorname{MSO}\to H\mathbb{Z}$$ does not split, but one can in fact work out explicit obstructions which lead to the examples you referred to. All this only depends on the homotopy type of $$X$$ (as opposed to the embedding question).
• @AchimKrause I can see why a splitting of $MSO\to H\mathbb{Z}$ would make the homology map surjective, but why does the failure of the splitting imply the failure of homology surjectivity? – Greg Friedman Oct 27 at 18:12
Turns out (as @archipelago noted) this is called the Steenrod Problem. The answer I linked also holds for $$X$$ that are not manifolds. In particular every class can be represented for $$n \leq 6$$. There are example for classes that cannot be represented for $$n = 7$$