Homotopy Groups are Homeomorphism Invariants

Theorem

Let $X$ and $Y$ be two manifolds.

Let there exist a homeomorphism $\phi: X \to Y$.

Then:

$\forall n \in \N: \pi_n (X) = \pi_n(Y)$

where the $\pi_n$ are the $n^{th}$ homotopy groups.

Proof

Let $\phi$ be any homeomorphism $\phi:X\to Y$. We must show that:

1. If $c:[0,1]^n \to X$ is a continuous mapping, then $ \phi \circ c:[0,1]^n \to Y$ is as well;
2. If $c,d:[0,1]^n \to X$ are homotopic, then $\phi\circ c, \phi\circ d:[0,1]^n \to Y$ are homotopic as well;
3. If $c,d:[0,1]^n \to X$ are not homotopic, there can be no homotopy between $\phi\circ c$ and $\phi \circ d$;
4. The image of the concatenation of two maps, $\phi(c * d)$, is the concatenation of the images, $\phi(c)*\phi(d)$.