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It seems that if I take any n points in a connected manifold of 2 dimensions or more, $M$, $x_i$, and any other points in $M$, $y_i$, I can continuously move every point in $M$ around until $x_1$ is at $y_1$, $x_2$ is at $y_2$, etc. I think this smushing is described by a homeomorphism homotopic to the identity map. This seems to be a very important property:

It allows us to define wedge products up to homeomorphism without specifying join points. It allows us to do many topological proofs that involve sliding one thing along another much more easily, by avoiding having to explicitly construct the function describing the sliding. I think it is necessary to prove the Euler characteristic of a manifold is constant, by placing the vertices of one graph embedded in $M$ in the same place as another graph in $M$.

I think that this property can be proved by topological induction on the product of n copies of $M$, because for any choice of $x_i$, you can find neighborhoods around them locally homeomorphic to $\mathbb{R}^n$ (making sure the neighborhoods don't overlap), and use that local homeomorphism to define a smooth map that takes the $x_i$ to any $y_i$ within these neighborhoods. This means there is an open neighborhood of $(x_1, ..., x_n)$ in $M^n$ that can be reached. So the set of points in $M^n$ that can be reached this way is open. Similarly, the complement is open.

However, all $n$ points must be distinct, so the space I'm working on is $M^n$ with some 1D manifolds removed, but this should still be connected. As $M^n-B$ is connected, every point must be reachable. I would also like to check that this proof strategy works.

Edit: Here is an example of how I think you could define the map.

Take one of the $x_i$ and an open neighborhood $A_i$ around it, which is homeomorphic to $\mathbb{R}^n$. Choose it also so that the closure of $A_i$ is homeomorphic to the closed ball. Then I can define a map from the closed ball to itself which leaves the boundary fixed, while mapping 0 to another point, $y$, in the ball. Take the function $$f(x) = x + (1 - |x|)y$$ for example. This shows I can move any point in the interior of the closed ball to any other point while leaving the boundary fixed. Applying this to $A_i$ allows me to move $x_i$ to any $y_i$ in $A_i$, while leaving every point outside of $A_i$ fixed. As the $A_i$ don't overlap, I can do this for each $x_i$ at once without them interacting.

Zoe Allen
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  • No, this doesn't work. How do you know your local homeo can define a smooth map taking $x_i$ to $y_i$? And if so, how would this interact with the other $x_j$? – Randall Nov 23 '23 at 14:19
  • @randall I am not assuming the manifolds are smooth, so I'm not requiring smooth maps. The idea is, for a given $x_i$ and neighborhood $A_i$ around it, to use the homeomorphism $A_i$ to $\mathbb{R}^n$ to define a continuous map which leaves every point not in $A_i$ where it is, while moving $x_i$ to some point $y_i$ in $A_i$. Because the other points are left alone, and the $A_i$ don't overlap, I can do these maps all at once without them interacting. – Zoe Allen Nov 23 '23 at 14:26
  • It is going to be hard to leave every point not in $A_i$ where it is and still have continuity. – Randall Nov 23 '23 at 14:30
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    @Randall I've editted the post to add how I think I can do that. – Zoe Allen Nov 23 '23 at 14:52
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    https://math.stackexchange.com/questions/135691/a-question-about-the-topological-homogeneity-of-manifolds?noredirect=1&lq=1 – Moishe Kohan Nov 25 '23 at 04:18
  • @MoisheKohan thank you, the additional note here is exactly what I'm asking, and confirms that this proof strategy works. Now I just need to know how to prove $M^n-$the diagonals is connected. They don't mention a name so I think that resolves my question. – Zoe Allen Nov 25 '23 at 15:08
  • Just do at the dimension count: What is the dimension of $M^n$ and what is the dimension of the "diagonals." – Moishe Kohan Nov 25 '23 at 15:09
  • @MoisheKohan intuitively it is clear to me that that works, but I don't know how to prove it. – Zoe Allen Nov 25 '23 at 15:10
  • Do you know how to prove that a linear subspace of codimension $\ge 2$ does not separate ${\mathbb R}^k$? How about a union of finitely many linear subspaces? – Moishe Kohan Nov 25 '23 at 15:12

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