A description/proof of that for 2 homotopic maps, point on domain, and fundamental group homomorphisms induced by maps, 2nd homomorphism is composition of canonical 'groups - group homomorphisms' isomorphism between codomains of homomorphisms after 1st homomorphism
Topics
About: topological space
About: group
The table of contents of this article
Starting Context
- The reader knows a definition of homotopic map.
- The reader knows a definition of fundamental group homomorphism induced by map.
- The reader knows a definition of %category name% isomorphism.
- The reader knows a definition of affine map.
- The reader admits the proposition that for any 2 path-connected points on any topological space, there is a 'groups - group homomorphisms' isomorphism between the fundamental groups with respect to the 2 points that (the isomorphism) multiplies the inverse-path class from the left and the path class from the right in the path classes groupoid.
- The reader admits the proposition that any map between topological spaces is continuous if the domain restriction of the map to each closed set of a finite closed cover is continuous.
Target Context
- The reader will have a description and a proof of the proposition that for any 2 homotopic maps, any point on the domain, and the fundamental group homomorphisms induced by the maps, the 2nd homomorphism is the composition of the canonical 'groups - group homomorphisms' isomorphism between the codomains of the homomorphisms after the 1st homomorphism.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Description
For any topological spaces, \(T_1, T_2\), any homotopic maps, \(f: T_1 \to T_2\) and \(f': T_1 \to T_2\), such that \(f \simeq f'\), any point, \(p \in T_1\), the fundamental group homomorphisms induced by \(f\) and \(f'\), \(f_*: \pi_1 (T_1, p) \to \pi_1 (T_2, f (p))\) and \(f'_*: \pi_1 (T_1, p) \to \pi_1 (T_2, f' (p))\), and the canonical 'groups - group homomorphisms' isomorphism, \(\phi: \pi_1 (T_2, f (p)) \to \pi_1 (T_2, f' (p))\), \(f'_* = \phi \circ f_*\).
2: Proof
There is a homotopy, \(F: T_1 \times I \to T_2\), such that \(F (p', 0) = f' (p')\) and \(F (p', 1) = f (p')\). Let \(\gamma: I \to T_2 = F (p, 1 - t)\), which is a path such that \(\gamma (0) = F (p, 1) = f (p)\) and \(\gamma (1) = F (p, 0) = f' (p)\).
\(\phi\) is indeed a 'groups - group homomorphisms' isomorphism and \(\phi: [g] \mapsto [\gamma^{-1}] [g] [\gamma]\) where \([\gamma^{-1}] [g] [\gamma]\) is in the path classes groupoid, by the proposition that for any 2 path-connected points on any topological space, there is a 'groups - group homomorphisms' isomorphism between the fundamental groups with respect to the 2 points that (the isomorphism) multiplies the inverse-path class from the left and the path class from the right in the path classes groupoid.
Let \([g] \in \pi_1 (T_1, p)\) be any. \(f'_* ([g]) = [f' \circ g]\). \(\phi \circ f_* ([g]) = \phi ([f \circ g]) = [\gamma^{-1}] [f \circ g] [\gamma] = [\gamma^{-1} * (f \circ g) * \gamma]\). \(f' \circ g \simeq \gamma^{-1} * (f \circ g) * \gamma \text{ rel } \{0, 1\}\)?
Let us define \(F': I \times I \to T_2\), \((t, s) \mapsto F (g (t), s)\). \(F'\) is continuous as a composition of continuous maps. \(F' (t, 0) = f' \circ g (t)\) as desired, but \(F' (t, 1) = f \circ g (t)\) as not desired. Besides, \(F' (0, s) = F (g (0), s)\) and \(F' (1, s) = F (g (1), s)\) as not desired.
Let us think of the square, \(S\), that represents \(I \times I\) where the bottom side is \(s = 0\), the top side is \(s = 1\), the left side is \(t = 0\), and the right side is \(t = 1\). In fact, \(\gamma^{-1} * (f \circ g) * \gamma\) corresponds to going \((0, 0) \to (0, 1) \to (1, 1) \to (1, 0)\).
So, let us contrive a continuous map, \(h: I \times I \to I \times I\), such that \((0, 1) \to (1, 1)\) maps to \((0, 0) \to (0, 1) \to (1, 1) \to (1, 0)\).
Let us think of the square, \(S'\), that represents the domain \(I \times I\) as before. \(h\) is a map from \(S'\) into \(S\). In order for the top side of \(S'\) to correspond to \(\gamma^{-1} * (f \circ g) * \gamma\), \((0, 1) \to (1 / 4, 1)\) has to correspond to \(\gamma^{-1}\), \((1 / 4, 1) \to (1 / 2, 1)\) has to correspond to \(f \circ g\), and \((1 / 2, 1) \to (1, 1)\) has to correspond to \(\gamma\). Let the center, \((1/ 2, 1 /2)\), of \(S'\) be denoted as \(c\). Let us divide \(S'\) into the triangles, \((0, 0) - c - (0, 1)\), \((0, 1) - c - (1 / 4, 1)\), \((1 / 4, 1) - c - (1 / 2, 1)\), \((1 / 2, 1) - c - (1, 1)\), \((1, 1) - c - (1, 0)\), and \((1, 0) - c - (0, 0)\). Let \(h\) map \(S'\) onto \(S\) by affine mapping each triangle onto a triangle on \(S\). Each affine map is defined by mappings of the vertices, which determines the whole map by the linearity.
This is the mappings of the vertices: \(c \to (1 / 2, 1 / 2)\), \((0, 0) \to (0, 0)\), \((0, 1) \to (0, 0)\), \((1 / 4, 1) \to (0, 1)\), \((1 / 2, 1) \to (1, 1)\), \((1, 1) \to (1, 0)\), \((1, 0) \to (1, 0)\). Then, \((0, 1) \to (1, 1)\) on \(S'\) maps to \((0, 0) - (0, 1) - (1, 1) - (1, 0)\) on \(S\) as desired. Besides, \((0, 0) \to (0, 1)\) maps to the single point, \((0, 0)\), and \((1, 0) \to (1, 1)\) maps to the single point, \((1, 0)\). That constitutes a consistent map (the maps on the triangles coincide on the borders), \(h\), because it does not map any vertex to 2 different points. Each triangle on \(S'\) is a closed subset of \(S'\), and \(h\) is continuous on the closed subset (because it is an affine map), so, \(h\) is continuous, by the proposition that any map between topological spaces is continuous if the domain restriction of the map to each closed set of a finite closed cover is continuous.
\(F'': I \times I \to T_2 = F' \circ h\) is a desired homotopy relative to \(\{0, 1\}\), because \(F'' (t, 0) = f' \circ g (t)\), \(F'' (t, 1) = (\gamma^{-1} * (f \circ g) * \gamma) (t)\), \(F'' (0, s) = f' \circ g (0) = (\gamma^{-1} * (f \circ g) * \gamma) (0) = f' (p) = F'' (1, s) = f' \circ g (1) = (\gamma^{-1} * (f \circ g) * \gamma) (1) = f' (p)\).
So, \(f' \circ g \simeq \gamma^{-1} * (f \circ g) * \gamma \text{ rel } \{0, 1\}\), and \([f' \circ g] = [\gamma^{-1} * (f \circ g) * \gamma]\), which means that \(f'_* = \phi \circ f_*\).
