419: Fundamental Group Homomorphism Induced by Homotopy Equivalence Is 'Groups - Group Homomorphisms' Isomorphism

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A description/proof of that fundamental group homomorphism induced by homotopy equivalence is 'groups - group homomorphisms' isomorphism

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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 fundamental group homomorphism induced by map.
• The reader knows a definition of homotopy equivalence.
• The reader knows a definition of %category name% isomorphism.
• The reader admits 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.
• The reader admits the proposition that any bijective group homomorphism is a 'groups - group homomorphisms' isomorphism.

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Target Context

• The reader will have a description and a proof of the proposition that the fundamental group homomorphism induced by any homotopy equivalence is a 'groups - group homomorphisms' isomorphism.

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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 homotopy equivalence, $f:T_1\to T_2$, and any point, $p\in T_1$, the fundamental group homomorphism induced by $f$, $f_{\ast }:{\pi }_1(T_1,p)\to {\pi }_1(T_2,f(p))$, is a 'groups - group homomorphisms' isomorphism.

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2: Proof

There is a continuous map, $f^{\prime }:T_2\to T_1$, such that $f^{\prime }\circ f\simeq 1_{T_1}$ and $f\circ f^{\prime }\simeq 1_{T_2}$.

There is a homotopy, $F:T_1\times I\to T_1$, such that $F(p^{\prime },0)=f^{\prime }\circ f(p^{\prime })$ and $F(p^{\prime },1)=1_{T_1}(p^{\prime })=p^{\prime }$. $(f^{\prime }\circ f{)}_{\ast }:{\pi }_1(T_1,p)\to {\pi }_1(T_1,f^{\prime }\circ f(p))$ and $(1_{T_1}{)}_{\ast }:{\pi }_1(T_1,p)\to {\pi }_1(T_1,p)$. There is the canonical 'groups - group homomorphisms' isomorphism, $\varphi :{\pi }_1(T_1,f^{\prime }\circ f(p))\to {\pi }_1(T_1,p)$, $[g]\mapsto [{\gamma }^{-1}][g][\gamma ]$, where $\gamma :I\to T_1$, $t\mapsto F(p,1-t)$, such that $(1_{T_1}{)}_{\ast }=\varphi \circ (f^{\prime }\circ f{)}_{\ast }$, by 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.

As $(1_{T_1}{)}_{\ast }$ is a bijection and $\varphi$ is a bijection, $(f^{\prime }\circ f{)}_{\ast }$ is a bijection. $(f^{\prime }\circ f{)}_{\ast }=f_{\ast }^{\prime }\circ f_{\ast }$, and as the left hand side is a bijection, $f_{\ast }$ is an injection and $f_{\ast }^{\prime }$ is a surjection.

By the likewise argument for $f\circ f^{\prime }\simeq 1_{T_2}$, $f_{\ast }^{\prime }$ is an injection and $f_{\ast }$ is a surjection.

So, $f_{\ast }$ is a bijection.

So, $f_{\ast }$ is a 'groups - group homomorphisms' isomorphism, by the proposition that any bijective group homomorphism is a 'groups - group homomorphisms' isomorphism.

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References

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