422: 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

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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

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Topics

About: topological space
About: group

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof

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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.

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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.

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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.

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Main Body

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1: Description

For any topological spaces, $T_1,T_2$, any homotopic maps, $f:T_1\to T_2$ and $f^{\prime }:T_1\to T_2$, such that $f\simeq f^{\prime }$, any point, $p\in T_1$, the fundamental group homomorphisms induced by $f$ and $f^{\prime }$, $f_{\ast }:{\pi }_1(T_1,p)\to {\pi }_1(T_2,f(p))$ and $f_{\ast }^{\prime }:{\pi }_1(T_1,p)\to {\pi }_1(T_2,f^{\prime }(p))$, and the canonical 'groups - group homomorphisms' isomorphism, $\varphi :{\pi }_1(T_2,f(p))\to {\pi }_1(T_2,f^{\prime }(p))$, $f_{\ast }^{\prime }=\varphi \circ f_{\ast }$.

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2: Proof

There is a homotopy, $F:T_1\times I\to T_2$, such that $F(p^{\prime },0)=f^{\prime }(p^{\prime })$ and $F(p^{\prime },1)=f(p^{\prime })$. Let $\gamma :I\to T_2=F(p,1-t)$, which is a path such that $\gamma (0)=F(p,1)=f(p)$ and $\gamma (1)=F(p,0)=f^{\prime }(p)$.

$\varphi$ is indeed a 'groups - group homomorphisms' isomorphism and $\varphi :[g]\mapsto [{\gamma }^{-1}][g][\gamma ]$ 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. $f_{\ast }^{\prime }([g])=[f^{\prime }\circ g]$. $\varphi \circ f_{\ast }([g])=\varphi ([f\circ g])=[{\gamma }^{-1}][f\circ g][\gamma ]=[{\gamma }^{-1}\ast (f\circ g)\ast \gamma ]$. $f^{\prime }\circ g\simeq {\gamma }^{-1}\ast (f\circ g)\ast \gamma \text{ rel }\{0,1\}$?

Let us define $F^{\prime }:I\times I\to T_2$, $(t,s)\mapsto F(g(t),s)$. $F^{\prime }$ is continuous as a composition of continuous maps. $F^{\prime }(t,0)=f^{\prime }\circ g(t)$ as desired, but $F^{\prime }(t,1)=f\circ g(t)$ as not desired. Besides, $F^{\prime }(0,s)=F(g(0),s)$ and $F^{\prime }(1,s)=F(g(1),s)$ 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, ${\gamma }^{-1}\ast (f\circ g)\ast \gamma$ 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, $S^{\prime }$, that represents the domain $I\times I$ as before. $h$ is a map from $S^{\prime }$ into $S$. In order for the top side of $S^{\prime }$ to correspond to ${\gamma }^{-1}\ast (f\circ g)\ast \gamma$, $(0,1)\to (1/4,1)$ 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 $S^{\prime }$ be denoted as $c$. Let us divide $S^{\prime }$ 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 $S^{\prime }$ 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 $S^{\prime }$ 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 $S^{\prime }$ is a closed subset of $S^{\prime }$, 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.

$F^{″}:I\times I\to T_2=F^{\prime }\circ h$ is a desired homotopy relative to $\{0,1\}$, because $F^{″}(t,0)=f^{\prime }\circ g(t)$, $F^{″}(t,1)=({\gamma }^{-1}\ast (f\circ g)\ast \gamma )(t)$, $F^{″}(0,s)=f^{\prime }\circ g(0)=({\gamma }^{-1}\ast (f\circ g)\ast \gamma )(0)=f^{\prime }(p)=F^{″}(1,s)=f^{\prime }\circ g(1)=({\gamma }^{-1}\ast (f\circ g)\ast \gamma )(1)=f^{\prime }(p)$.

So, $f^{\prime }\circ g\simeq {\gamma }^{-1}\ast (f\circ g)\ast \gamma \text{ rel }\{0,1\}$, and $[f^{\prime }\circ g]=[{\gamma }^{-1}\ast (f\circ g)\ast \gamma ]$, which means that $f_{\ast }^{\prime }=\varphi \circ f_{\ast }$.

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References

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