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To be clear, I know these sets are not diffeomorphic or even homeomorphic in general. However, I've been told that there doesn't even exist a bijection between these sets.

But suppose $M$ is an $n$-dimensional manifold and let $\{\partial_1|_p, \ldots, \partial_n|_p\}$ be the basis of $T_p M$ with respect to some chart containing $p \in M$. If $v_p \in T_p M$ we have $v_p = v_p^i \partial_i|_p$ for unique real numbers $v_p^i$. Define the function $\lambda: TM \to M \times \mathbb{R}^n$ by $\lambda(p, v_p)=(p, v_p^1, \ldots, v_p^n)$.

Surely this is a well-defined bijection?

Blonge
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  • Yes, you have well-defined bijection. Perhaps the claim was that there is no natural bijection? As you have demonstrated, it's possible to write a bijection once you choose a basis for each tangent space $T_pM$. – Brian Shin Oct 06 '20 at 00:14

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Yes, you are correct that you can get a bijection by picking a basis for each tangent space. Perhaps whoever claimed there was no such bijection meant that there was no canonical bijection (and so would exclude yours since it involves an arbitrary choice of basis for each tangent space).

(I would also remark that the existence of a bijection is an extremely weak condition. Indeed, every positive-dimensional nonempty (second countable) manifold has cardinality $2^{\aleph_0}$, so there are bijections between all of them.)

Eric Wofsey
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    This bijection has the special property that it restricts to linear isomorphisms on the tangent spaces $T_p M$. So it's a little better than just picking a random bijection! – Blonge Oct 06 '20 at 00:46
  • How can I verify your last remark about cardinality? – Kelvin Lois Oct 06 '20 at 06:12
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Sure: The tangent bundle on $M^n$ is by definition the disjoint union (with a more complicated topology) of the $T_p M$ for $p\in M$, and each $T_p M$ is an $n$-dimensional vector space by definition. Of course, the isomorphisms $T_p M \to \mathbb{R}^n$ are not canonical and can't be made continuous in $p$ for a nontrivial bundle, but that's not a problem if you're just looking for a map of sets.

anomaly
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