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dg.differential geometry – Do smooth maps with nowhere-maximal rank have small image?

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 $f: Wto M$ has small image if there is another smooth map of manifolds $g: Tto M$ 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!

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Topological stable rank of $C_0(mathbb{Z} times K)$

I was trying to calculate the topological stable rank of $C_0(mathbb{Z} times K)$ where $K$ is Cantor set. My idea: Since $C_0(mathbb{Z} times K)$ is not unital if we some how make it unital then $spec (C_0(mathbb{Z} times K))$ is homeomorphic to Cantor set and $C_0(mathbb{Z} times K)$ became AF-algebra. but I can’t make the algebra unital. Does anybody have any idea to make $C_0(mathbb{Z} times K)$ unital?

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