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.