Homotopy

Definition homotopy

• $X$,
  • top sp
• $I := [0, 1]$,
• $f: I \rightarrow X$,
  • continuous map
• $g: I \rightarrow X$,
  • continuous map
• $H: I \times I \rightarrow X$,
  • continuous map

$H$ is sait to be homotopy between $f$ and $g$.

\[H(0, t) := f(t), \ H(1, t) := g(t).\]

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Definition homotop

• $X$,
  • top sp
• $I := [0, 1]$,
• $f: I \rightarrow X$,
  • continuous map
• $g: I \rightarrow X$,
  • continuous map

$f$ and $g$ is said to be homotop if there exists homotopy $H$ between $f$ and $h$. We write $f \cong g$.

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Remark

Homotop is a equivalence relation. Let $f \cong \cong h$ and $H_{1}$ and $H_{2}$ be homotopy of $f \cong g$ and $g \cong h$ respecitvely. We’ll show that $f \cong h$. Let

\[H(s, t) := \begin{cases} H(2s, t) & (0 \le s \le 1/2) \\ H(2s - 1, t) & (1/2 \le s \le 1) \end{cases} .\]

$H$ is a homotopy between $f$ and $h$. Reflective and symmetric properties are obvious.

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Definition

• $f:I \rightarrow X$,
  • path

\[[f] := \{ g: \mathrm{pat} \mid f \cong g \} .\] \[f^{-1}(t) := f(1 - t) .\] \[0_{x}(t) \equiv x .\]

For any $x, y \in X$, a set of path from $x$ to $y$ $P(X; x, y)$ is defind

\[P(X; x, y) := \{ f: I \rightarrow X: \mathrm{continueous map} \mid f(0) = x, \ f(1) = y \} .\]

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Definition fundamental group

• $X$,
  • top sp.
• $x \in X$,

Let

\[\pi(x, X) := P(X; x, x) / \cong .\]

We define multiplication is defined

\[[f] [g] := [fg] .\]

The inverse is $[f]^{-1} := [f^{-1}]$. And unit is $[0_{x}]$.

$pi(x, X)$ is called a fundamental group.

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Remark

$pi(x, X)$ is a group.

Indeed, $[f][g]$ is well-defined. Let $f^{\prime}$ and $g^{\prime}$ be another element in $[f]$ and $[g]$ respecitvely. We’ll show that $[fg] = [f^{\prime}g^{\prime}]$. There eixst homotopies

\[\begin{eqnarray} H_{f}(0, t) & = & f(t) \nonumber \\ H_{f}(1, t) & = & f^{\prime}(t) H_{g}(0, t) & = & g(t) \nonumber \\ H_{g}(1, t) & = & g^{\prime}(t) \nonumber . \end{eqnarray}\]

Let $H$ be

\[H(s, t) := \begin{cases} H(s, 2t) & (0 \le t \le 1/2) \\ H(s, 2t - 1) & (1/2 \le t \le 1) \end{cases} .\]

$H$ is homotopy betwee $f * g$ and $f^{\prime} * g^{\prime}$.

\([0_{x}]\) is unit. We’ll show that $[0_{x}][f] = [f]$. Let

\[H(s, t) := \begin{cases} x & (0 \le t \le (-s + 1) / 2) \\ f((2t - s - 1) / (s + 1)) & ( (-s + 1) / 2 \le t \le 1) \end{cases} .\]

Particularly,

\[\begin{eqnarray} H(0, t) & = & \begin{cases} x & (0 \le t \le 1 / 2) \\ f(2t - 1) & (1/ 2 \le t \le 1) \end{cases} \nonumber \\ H(1, t) & = & f(t) \ (0 \le t \le 1) \end{eqnarray}\]

$H$ is homotopy between $0_{x} * f$ and $f$.

Lastly, we’ll show that $[f] [f^{-1}] = [0_{x}]$. Let

\[H(s, t) := \begin{cases} f(2t) & (0 \le t \le s/2) \\ f(s) & (s/2 \le t \le 1 - t/2) \\ f(2 - 2s) & (1 - s/2 \le t \le 1) \end{cases}\] \[\begin{eqnarray} H(0, t) & = & \begin{cases} f(2t) & (0 \le t \le 0) \\ f(0) & (0 \le t \le 1) \end{cases} \nonumber \\ H(0, t) & = & \begin{cases} f(2t) & (0 \le t \le 0) \\ f(0) & (0 \le t \le 1) \end{cases} \end{eqnarray}\]

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Reference

• http://www.math.tsukuba.ac.jp/~tasaki/lecture/ln2019/diffgeoI2019-dist.pdf