Here is the situation I am concerned with -(adsbygoogle = window.adsbygoogle || []).push({});

Consider a smooth curve [itex]g:[0,1] \to M[/itex] where [itex]M [/itex] is a topological manifold (I'd be happy to assume [itex] M [/itex] smooth/finite dimensional if that helps). Let [itex] Im(g) [/itex] be the image of [itex] [0,1] [/itex] under the map [itex] g [/itex]. Give [itex] Im(g) [/itex] the subspace topology induced by [itex] M [/itex].

The question is this --- as a topological space, does [itex] Im(g) [/itex] always have (topological) dimension [itex] \leq 1 [/itex] ?

Note that it IS important not to restrict [itex] g [/itex] to be either injective or an immersion - then the result is straightforward. The tricky thing is that [itex] g [/itex] may not be constant rank and also may not be a submanifold of [itex] M [/itex] (or even a manifold at all).

Also, a reference to a proof is fine, I don't really need to know HOW to prove it, I just need to be certain that it is true. It certainly seems intuitively obvious...

I've seen a number of statements (without reference or proof) that [itex] Im(g) [/[STRIKE][/STRIKE]itex] must have zero Lebesque measure. I don't immediately see how this would answer the above question, so any references or proofs on this front would be useful as well.

Thanks!

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# Topological dimension of the image of a smooth curve in a manifold

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