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By general manifold I mean Hausdorff differential manifold not necessarily second-countable. By standard manifold I mean Hausdorff, second-countable differential manifold.

So my question is, we have an injective immersion, ie with a non singular derivative, $\iota\colon M\to N$. Where $M$ is a connected general manifold and $N$ is a standard manifold. Is it true that $M$ is second-countable.

Note that if we take away the connection hypothesis then it is false. If we consider $M$ to be $(0,1)\times (0,1)$ with the horizontal topology, ie open basis of the form ${a}\times U$ with U open set of $(0,1)$; $N$ to be $(0,1)\times (0,1)$ with the standard topology, and $\iota$ to be the identity then it is false.

If we take away the injective hypothesis I think it is also false. I'm not really sure if it can be done, but if it is possible to wind up the long line around $\mathbb S^1$ similar to how we do with $\mathbb R$ it would be a counterexample.

I've been told it can be done with Riemann manifolds. Something like: $N$ has a riemann metric because it is second-countable, $\iota(M)$ has a riemann metric induced by $N$, taking it back to $M$ by $\iota$, you get a metric in $M$ so it has to be second-countable. However I don't know anything about Riemann Geometry so I'd prefer a pure topological/differential proof.

I'm interested because it allows you in the proof of the Frobenius theorem not to check about second-countable in the manifold you obtain.

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1 Answer 1

up vote 8 down vote accepted

Warm-up
In contrast to the winding of $\mathbb R$, it is impossible to immerse the long line $L$ into the circle $S^1$.
Indeed if you had such an immersion (injective or not) $i:L\to S^1$, it would induce an isomorphism of line bundles $T_L \stackrel {\simeq}{\to} i^*T_S$ on $L$.
Since the tangent bundle of $S^1$ is trivial, the same would be true of $i^*T_S$ and then of $T_L$ .
However Morrow proved that $T_L$ is not trivial.

The general case
A non-second-countable manifold $M$ cannot be immersed into a second countable manifold $N$.
Indeed, if $i:M\to N$ is such an immersion, put a Riemannian structure on the tangent bundle $T(N)$, which is possible since $N$ is paracompact.
The immersion gives rise to an embedding $di:T(M)\hookrightarrow i^*T(N)$ of vector bundles on $M$, which induces a Riemannian structure on $T(M)$. This Riemannian structure in turn endows $M$ with a metric compatible with its topolgy. However a connected metrizable manifold is second-countable . Contradiction.
(I have used that second-countability of a connected manifold is equivalent to metrizability or paracompactness: see here)

Edit: An alternative proof
In case $M$ and $N$ have same dimension one can give a purely topological proof, more in accordance with your wish of avoiding Riemannian structures.
(It suffices to give a riemannless proof of the warm-up) . Here goes:

Recall that the Poincaré-Volterra theorem states that if $i:X\to Y$ is a local homeomorphism with $X$ connected Hausdorff and $Y$ locally compact, locally connected and second-countable, then $X$ is second-countable too.
If $M$ and $N$ have the same dimension , an immersion $i:M\to N$ is a local diffeomorphism and a fortiori a local homeomorphism. The theorem of Poincaré-Volterra thus proves that it is impossible to have both $M$ not second- countable and $N$ second-countable.

[Poincaré-Volterra's theorem is the very last result in the first chapter of Bourbaki's Topology.
It is used, for example, in the proof of Radó's theorem according to which every Riemann surface is second-countable.]

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Morrow's article is fabulous! I wasn't aware of this (surprisingly easy) result. Thanks for sharing it. –  Jason DeVito Jan 27 '12 at 20:12
    
Thanks, do you have any idea about the main question? –  Gabriel Furstenheim Jan 27 '12 at 20:55
    
Dear @Gabriel, well, yes: it is the same as what you were told . I have written down a complete proof in an edit and given a purely topological proof in the equidimensional case. –  Georges Elencwajg Jan 27 '12 at 21:49

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