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I’m trying to better understand the concept of “maps with small image” as used by Lipyanskiy in his construction of “geometric homology” in https://arxiv.org/abs/1409.1121. Lipyanskiy utilizes manifolds with corners, but for the purposes of this question I think it suffices to stick to ordinary manifolds, which we assume to be second countable.

By definition, a smooth map of manifolds $f: W\to M$ has small image if there is another smooth map of manifolds $g: T\to M$ such that $\dim(T)<\dim(W)$ and $f(W)\subset g(T)$. I’m interested in alternative formulations of this condition. In particular, it seems reasonable to conjecture that this condition is equivalent to the map $f$ having less than full rank at all points.

In fact, I’m pretty sure that having small image implies that $f$ is nowhere of full rank: otherwise $f$ will be an immersion at some point and so the image will have dimension at least $\dim(W)$. Then I believe the argument about Hausdorff dimension from this question about space filling curves implies that $f(W)$ cannot be covered by a smooth map with domain of smaller dimension: Proof that no differentiable space-filling curve exists

So my question really comes down to the converse: if $f$ nowhere achieves maximal rank, does it have small image?

I would also be interested in any other equivalent conditions to having small image.

Thanks!

Greg Friedman
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    Shouldn't the same argument work for that? If the rank is nowhere full, there is an open subset where the rank is constant $< \dim W$, and there the rank theorem shows that locally the map factors through a lower-dimensional thing. Now you're missing a closed subset where the rank is even smaller, but you should be able to stratify to break the problem up into pieces of constant rank. – R. van Dobben de Bruyn Jan 12 '21 at 21:18
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    @R.vanDobbendeBruyn Why is $W$ a union of locally closed submanifolds for which the rank is constant? I seem to need that to be able to run the end of your argument. (This reminds me of a question in a similar spirit which I do not know how to do: show that if $df_p$ is nowhere full rank, then $f$ is nowhere locally injective using purely smooth techniques, no homology. I run into trouble at these lower 'strata'...) – mme Jan 12 '21 at 22:17
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    It seems to me the question is local, and may be reduced to: suppose $f: \Bbb R^n \to \Bbb R^m$ is smooth, $n \leq m$, and $f(0) = 0$ with $df_0$ of less than full rank. Then $\text{Im}(f) \cap B_\epsilon(0)$ may be covered by the smooth image of countably many copies of $\Bbb R^{n-1}$. I don't know how to do this, though, and my impression is that singularity theory is not suitable for this non-generic question. One thing I haven't thought about: can you simply replace Lipyanskiy's condition with yours and get a homology theory again? – mme Jan 12 '21 at 22:25
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    To illustrate the point, I think you can have wild smooth maps $S^2 \to \Bbb R^2$ such that the "rank stratification" is far from a stratification. Consider Figure 1 in the notes here; this is a TOP embedding $S^2 \to \Bbb R^3$ and can be made to be a smooth map (derivative will vanish on the bad Cantor set points) by a "slowdown trick". Consider projecting this to the plane $\Bbb R^2$ of the paper by collapsing each "tube" to an arc ($S^1 \times I \to I$). This has rank <=1, "rank 0 stratum" a Cantor set on the sphere. – Balarka Sen Jan 12 '21 at 23:11
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    In fact it's not clear to me that the image of this map is contained in image of a map from a smooth $1$-manifold to the plane at all. Maybe one can make things worse by repositioning the Cantor set of non-flat points in the Alexander horned sphere to be vertical to the plane of the paper, in which case you'll have collapsed the Cantor set to a point. After thinking idly about these bad maps, my inclination is that the conjecture is false. – Balarka Sen Jan 12 '21 at 23:15
  • @R.vanDobbendeBruyn As Mike says, I found a paper of Boardman's ("Singularities of differentiable maps") which seems to suggest that earlier work of Thom showed that the generic situation provides a stratification by rank like you're talking about, but apparently this doesn't hold in general. – Greg Friedman Jan 14 '21 at 00:49
  • @MikeMiller Yes, we're thinking about replacing the small image condition with the small rank condition, but before doing that we wanted to have a better sense of whether or not these might actually be the same condition. – Greg Friedman Jan 14 '21 at 00:50

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