Homotopy Group is Homeomorphism Invariant

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Theorem

Let $X$ and $Y$ be two topological spaces.

Let $\phi: X \to Y$ be a homeomorphism.

Let $x_0 \in X$, $y_0 \in Y$.


Then for all $n \in \N$ the induced mapping:

$\phi_* : \pi_n \left({X, x_0}\right) \to \pi_n \left({Y, y_0}\right):$
$\left[{\!\left[{\, c \,}\right]\!}\right] \mapsto \left[{\!\left[{\, \phi \circ c \,}\right]\!}\right]$

is an isomorphism, where $\pi_n$ denotes the $n$th homotopy group.


Proof

Let $\phi: X \to Y$ be a homeomorphism.

We must show that:

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