Property of free group from categorical definition

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I'm proving some properties about the free groups using the categorical definition, which is the following:




Let $X$ be a set, $L$ a group and $i:Xto L$ a map. The pair $(L,i)$ is free on $X$ if for every group $H$ and every map $f:Xto H$ there exists a unique group homomorphism $varphi:Gto H$ such that $varphicirc i=f$.




I've already proved that



(1) $i$ must be injective and that



(2) if $(L_1,i_2)$ and $(L_2,i_2)$ are free on $X$ then there exists $varphi:L_1to L_2$ such that $varphicirc i_1=i_2$.



Now I'm stuck with the following:




If $|X|=|Y|$ and $(L_1,i_1)$ and $(L_2,i_2)$ are free on $X$ and $Y$ respectively, then $L_1cong L_2$.




I tried using a biyection $g:Xto Y$ and (2) to find $varphi:L_1to L_2$ which I can prove that is surjective. However, I cannot show that it is inyective outside $i_1(X)$, what can I do?




abstract-algebra group-theory category-theory free-groups




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  • I'm going to presume something along the lines of 'Note that there must be a similar $phi : L_2mapsto L_1$ and prove their compositions $phicircvarphi$ and $varphicircphi$ are natural isomorphisms?'
    – Steven Stadnicki
    Jul 31 at 17:38















up vote
2
down vote

favorite












I'm proving some properties about the free groups using the categorical definition, which is the following:




Let $X$ be a set, $L$ a group and $i:Xto L$ a map. The pair $(L,i)$ is free on $X$ if for every group $H$ and every map $f:Xto H$ there exists a unique group homomorphism $varphi:Gto H$ such that $varphicirc i=f$.




I've already proved that



(1) $i$ must be injective and that



(2) if $(L_1,i_2)$ and $(L_2,i_2)$ are free on $X$ then there exists $varphi:L_1to L_2$ such that $varphicirc i_1=i_2$.



Now I'm stuck with the following:




If $|X|=|Y|$ and $(L_1,i_1)$ and $(L_2,i_2)$ are free on $X$ and $Y$ respectively, then $L_1cong L_2$.




I tried using a biyection $g:Xto Y$ and (2) to find $varphi:L_1to L_2$ which I can prove that is surjective. However, I cannot show that it is inyective outside $i_1(X)$, what can I do?




abstract-algebra group-theory category-theory free-groups




share|cite|improve this question



















  • I'm going to presume something along the lines of 'Note that there must be a similar $phi : L_2mapsto L_1$ and prove their compositions $phicircvarphi$ and $varphicircphi$ are natural isomorphisms?'
    – Steven Stadnicki
    Jul 31 at 17:38













up vote
2
down vote

favorite









up vote
2
down vote

favorite











I'm proving some properties about the free groups using the categorical definition, which is the following:




Let $X$ be a set, $L$ a group and $i:Xto L$ a map. The pair $(L,i)$ is free on $X$ if for every group $H$ and every map $f:Xto H$ there exists a unique group homomorphism $varphi:Gto H$ such that $varphicirc i=f$.




I've already proved that



(1) $i$ must be injective and that



(2) if $(L_1,i_2)$ and $(L_2,i_2)$ are free on $X$ then there exists $varphi:L_1to L_2$ such that $varphicirc i_1=i_2$.



Now I'm stuck with the following:




If $|X|=|Y|$ and $(L_1,i_1)$ and $(L_2,i_2)$ are free on $X$ and $Y$ respectively, then $L_1cong L_2$.




I tried using a biyection $g:Xto Y$ and (2) to find $varphi:L_1to L_2$ which I can prove that is surjective. However, I cannot show that it is inyective outside $i_1(X)$, what can I do?




abstract-algebra group-theory category-theory free-groups




share|cite|improve this question











I'm proving some properties about the free groups using the categorical definition, which is the following:




Let $X$ be a set, $L$ a group and $i:Xto L$ a map. The pair $(L,i)$ is free on $X$ if for every group $H$ and every map $f:Xto H$ there exists a unique group homomorphism $varphi:Gto H$ such that $varphicirc i=f$.




I've already proved that



(1) $i$ must be injective and that



(2) if $(L_1,i_2)$ and $(L_2,i_2)$ are free on $X$ then there exists $varphi:L_1to L_2$ such that $varphicirc i_1=i_2$.



Now I'm stuck with the following:




If $|X|=|Y|$ and $(L_1,i_1)$ and $(L_2,i_2)$ are free on $X$ and $Y$ respectively, then $L_1cong L_2$.




I tried using a biyection $g:Xto Y$ and (2) to find $varphi:L_1to L_2$ which I can prove that is surjective. However, I cannot show that it is inyective outside $i_1(X)$, what can I do?





abstract-algebra group-theory category-theory free-groups





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share|cite|improve this question




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asked Jul 31 at 17:32









Javi

2,1481725




2,1481725











  • I'm going to presume something along the lines of 'Note that there must be a similar $phi : L_2mapsto L_1$ and prove their compositions $phicircvarphi$ and $varphicircphi$ are natural isomorphisms?'
    – Steven Stadnicki
    Jul 31 at 17:38

















  • I'm going to presume something along the lines of 'Note that there must be a similar $phi : L_2mapsto L_1$ and prove their compositions $phicircvarphi$ and $varphicircphi$ are natural isomorphisms?'
    – Steven Stadnicki
    Jul 31 at 17:38
















I'm going to presume something along the lines of 'Note that there must be a similar $phi : L_2mapsto L_1$ and prove their compositions $phicircvarphi$ and $varphicircphi$ are natural isomorphisms?'
– Steven Stadnicki
Jul 31 at 17:38





I'm going to presume something along the lines of 'Note that there must be a similar $phi : L_2mapsto L_1$ and prove their compositions $phicircvarphi$ and $varphicircphi$ are natural isomorphisms?'
– Steven Stadnicki
Jul 31 at 17:38











2 Answers
2






active

oldest

votes

















up vote
2
down vote



accepted










Fix a bijection $g: Xrightarrow Y$ with inverse $h: Yrightarrow X$.
Then $f_1= i_2circ g: Xrightarrow L_2$ defines a $varphi_1: L_1rightarrow L_2$ and similarly $f_2= i_1circ h: Yrightarrow L_1$ defines a $varphi_2: L_2rightarrow L_1$.
The composition $varphi_2circ varphi_1: L_1rightarrow L_1$ is the homomorphisms corresponding to the mapping $id_X: Xrightarrow X$. As this map "extends" uniquely to a homomorphism $L_1rightarrow L_1$ and the identity $id_L_1$ is obviously such a homomorphism, we have $varphi_2circ varphi_1= id_L_1$. Similarly $varphi_1circ varphi_2= id_L_2$, thus $varphi_1$ is the desired isomorphism.






share|cite|improve this answer





















  • To "extend" the map to a homomorphism $L_1to L_1$ do I have to assume that $L_1=langleXrangle$? Something that I haven't proved yet. Otherwise I don't know how to find such homomorphism using the definition of free group.
    – Javi
    Jul 31 at 18:14











  • No, it makes no difference if $X$ is contained in $L_1$ or not. I used the word extend in a loose way.
    – A. Pongrácz
    Jul 31 at 18:18






  • 1




    I think you meant: "The composition $φ2∘φ1:L_1→L_1$ is the homomorphism corresponding to the mapping $i:X→L$." Since free extensions only work on maps starting at $X$ and ending at some group, which necessarily cannot be the set $X$.
    – Musa Al-hassy
    Aug 1 at 15:12

















up vote
1
down vote













While there is already an accepted answer, I'd like to showcase a more algebraic solution to the problem which is not as common.



That $L, 𝒾$ is the free group for $X$ means that there is a way to
“extend” maps from $X$ to homomorphisms from $L$,



$$ ⟨_⟩ ,:, (X → H) ,⟶, (L → H) $$



Such that this is indeed an extension when restricted to the elements of $X$,



$$ (0) qquad ϕ ∘ 𝒾 = f quad≡quad ϕ = ⟨f⟩ $$



In (0) taking $ϕ$ to be $⟨f⟩$ yields



$$ (1) qquadqquadqquadqquad ⟨f⟩ ∘ 𝒾 = f $$



In (0) taking $ϕ, f$ to be $id, i$ yields



$$ (2) qquadqquadqquad;;;;qquad id = ⟨𝒾⟩ $$



Now suppose we are given $g : X ≅ Y$ where
$X, Y$ have free groups $L₁, 𝒾₁$ and $L₂, 𝒾₂$, then we
claim $⟨i₂ ∘ g⟩ : L₁ ≅ L₂$ with inverse $⟨𝒾₁ ∘ g⁻¹⟩$.



Indeed we can show this to be true with a simple calculation:
--omitting the subscripts--



 ⟨𝒾 ∘ g⟩ ∘ ⟨𝒾 ∘ g⁻¹⟩ = id
≡ Using (2) with an aim to utilising (0) 
 ⟨𝒾 ∘ g⟩ ∘ ⟨𝒾 ∘ g⁻¹⟩ = ⟨𝒾⟩
≡ Now in a position to utilise (0) 
 ⟨𝒾 ∘ g⟩ ∘ ⟨𝒾 ∘ g⁻¹⟩ ∘ 𝒾 = 𝒾
≡⟨ The only thing we can use is (1), so let's try it }
 ⟨𝒾 ∘ g⟩ ∘ 𝒾 ∘ g⁻¹ = 𝒾
≡⟨ Again the only thing we can use is (1), so let's try it }
 𝒾 ∘ g ∘ g⁻¹ = 𝒾
≡ Inverses and identity laws 
 𝒾 = 𝒾
≡ Reflexivity of equality 
 true





share|cite|improve this answer





















  • Thanks for your approach :)
    – Javi
    Aug 1 at 19:30










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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










Fix a bijection $g: Xrightarrow Y$ with inverse $h: Yrightarrow X$.
Then $f_1= i_2circ g: Xrightarrow L_2$ defines a $varphi_1: L_1rightarrow L_2$ and similarly $f_2= i_1circ h: Yrightarrow L_1$ defines a $varphi_2: L_2rightarrow L_1$.
The composition $varphi_2circ varphi_1: L_1rightarrow L_1$ is the homomorphisms corresponding to the mapping $id_X: Xrightarrow X$. As this map "extends" uniquely to a homomorphism $L_1rightarrow L_1$ and the identity $id_L_1$ is obviously such a homomorphism, we have $varphi_2circ varphi_1= id_L_1$. Similarly $varphi_1circ varphi_2= id_L_2$, thus $varphi_1$ is the desired isomorphism.






share|cite|improve this answer





















  • To "extend" the map to a homomorphism $L_1to L_1$ do I have to assume that $L_1=langleXrangle$? Something that I haven't proved yet. Otherwise I don't know how to find such homomorphism using the definition of free group.
    – Javi
    Jul 31 at 18:14











  • No, it makes no difference if $X$ is contained in $L_1$ or not. I used the word extend in a loose way.
    – A. Pongrácz
    Jul 31 at 18:18






  • 1




    I think you meant: "The composition $φ2∘φ1:L_1→L_1$ is the homomorphism corresponding to the mapping $i:X→L$." Since free extensions only work on maps starting at $X$ and ending at some group, which necessarily cannot be the set $X$.
    – Musa Al-hassy
    Aug 1 at 15:12














up vote
2
down vote



accepted










Fix a bijection $g: Xrightarrow Y$ with inverse $h: Yrightarrow X$.
Then $f_1= i_2circ g: Xrightarrow L_2$ defines a $varphi_1: L_1rightarrow L_2$ and similarly $f_2= i_1circ h: Yrightarrow L_1$ defines a $varphi_2: L_2rightarrow L_1$.
The composition $varphi_2circ varphi_1: L_1rightarrow L_1$ is the homomorphisms corresponding to the mapping $id_X: Xrightarrow X$. As this map "extends" uniquely to a homomorphism $L_1rightarrow L_1$ and the identity $id_L_1$ is obviously such a homomorphism, we have $varphi_2circ varphi_1= id_L_1$. Similarly $varphi_1circ varphi_2= id_L_2$, thus $varphi_1$ is the desired isomorphism.






share|cite|improve this answer





















  • To "extend" the map to a homomorphism $L_1to L_1$ do I have to assume that $L_1=langleXrangle$? Something that I haven't proved yet. Otherwise I don't know how to find such homomorphism using the definition of free group.
    – Javi
    Jul 31 at 18:14











  • No, it makes no difference if $X$ is contained in $L_1$ or not. I used the word extend in a loose way.
    – A. Pongrácz
    Jul 31 at 18:18






  • 1




    I think you meant: "The composition $φ2∘φ1:L_1→L_1$ is the homomorphism corresponding to the mapping $i:X→L$." Since free extensions only work on maps starting at $X$ and ending at some group, which necessarily cannot be the set $X$.
    – Musa Al-hassy
    Aug 1 at 15:12












up vote
2
down vote



accepted







up vote
2
down vote



accepted






Fix a bijection $g: Xrightarrow Y$ with inverse $h: Yrightarrow X$.
Then $f_1= i_2circ g: Xrightarrow L_2$ defines a $varphi_1: L_1rightarrow L_2$ and similarly $f_2= i_1circ h: Yrightarrow L_1$ defines a $varphi_2: L_2rightarrow L_1$.
The composition $varphi_2circ varphi_1: L_1rightarrow L_1$ is the homomorphisms corresponding to the mapping $id_X: Xrightarrow X$. As this map "extends" uniquely to a homomorphism $L_1rightarrow L_1$ and the identity $id_L_1$ is obviously such a homomorphism, we have $varphi_2circ varphi_1= id_L_1$. Similarly $varphi_1circ varphi_2= id_L_2$, thus $varphi_1$ is the desired isomorphism.






share|cite|improve this answer













Fix a bijection $g: Xrightarrow Y$ with inverse $h: Yrightarrow X$.
Then $f_1= i_2circ g: Xrightarrow L_2$ defines a $varphi_1: L_1rightarrow L_2$ and similarly $f_2= i_1circ h: Yrightarrow L_1$ defines a $varphi_2: L_2rightarrow L_1$.
The composition $varphi_2circ varphi_1: L_1rightarrow L_1$ is the homomorphisms corresponding to the mapping $id_X: Xrightarrow X$. As this map "extends" uniquely to a homomorphism $L_1rightarrow L_1$ and the identity $id_L_1$ is obviously such a homomorphism, we have $varphi_2circ varphi_1= id_L_1$. Similarly $varphi_1circ varphi_2= id_L_2$, thus $varphi_1$ is the desired isomorphism.







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 31 at 18:07









A. Pongrácz

1,314115




1,314115











  • To "extend" the map to a homomorphism $L_1to L_1$ do I have to assume that $L_1=langleXrangle$? Something that I haven't proved yet. Otherwise I don't know how to find such homomorphism using the definition of free group.
    – Javi
    Jul 31 at 18:14











  • No, it makes no difference if $X$ is contained in $L_1$ or not. I used the word extend in a loose way.
    – A. Pongrácz
    Jul 31 at 18:18






  • 1




    I think you meant: "The composition $φ2∘φ1:L_1→L_1$ is the homomorphism corresponding to the mapping $i:X→L$." Since free extensions only work on maps starting at $X$ and ending at some group, which necessarily cannot be the set $X$.
    – Musa Al-hassy
    Aug 1 at 15:12
















  • To "extend" the map to a homomorphism $L_1to L_1$ do I have to assume that $L_1=langleXrangle$? Something that I haven't proved yet. Otherwise I don't know how to find such homomorphism using the definition of free group.
    – Javi
    Jul 31 at 18:14











  • No, it makes no difference if $X$ is contained in $L_1$ or not. I used the word extend in a loose way.
    – A. Pongrácz
    Jul 31 at 18:18






  • 1




    I think you meant: "The composition $φ2∘φ1:L_1→L_1$ is the homomorphism corresponding to the mapping $i:X→L$." Since free extensions only work on maps starting at $X$ and ending at some group, which necessarily cannot be the set $X$.
    – Musa Al-hassy
    Aug 1 at 15:12















To "extend" the map to a homomorphism $L_1to L_1$ do I have to assume that $L_1=langleXrangle$? Something that I haven't proved yet. Otherwise I don't know how to find such homomorphism using the definition of free group.
– Javi
Jul 31 at 18:14





To "extend" the map to a homomorphism $L_1to L_1$ do I have to assume that $L_1=langleXrangle$? Something that I haven't proved yet. Otherwise I don't know how to find such homomorphism using the definition of free group.
– Javi
Jul 31 at 18:14













No, it makes no difference if $X$ is contained in $L_1$ or not. I used the word extend in a loose way.
– A. Pongrácz
Jul 31 at 18:18




No, it makes no difference if $X$ is contained in $L_1$ or not. I used the word extend in a loose way.
– A. Pongrácz
Jul 31 at 18:18




1




1




I think you meant: "The composition $φ2∘φ1:L_1→L_1$ is the homomorphism corresponding to the mapping $i:X→L$." Since free extensions only work on maps starting at $X$ and ending at some group, which necessarily cannot be the set $X$.
– Musa Al-hassy
Aug 1 at 15:12




I think you meant: "The composition $φ2∘φ1:L_1→L_1$ is the homomorphism corresponding to the mapping $i:X→L$." Since free extensions only work on maps starting at $X$ and ending at some group, which necessarily cannot be the set $X$.
– Musa Al-hassy
Aug 1 at 15:12










up vote
1
down vote













While there is already an accepted answer, I'd like to showcase a more algebraic solution to the problem which is not as common.



That $L, 𝒾$ is the free group for $X$ means that there is a way to
“extend” maps from $X$ to homomorphisms from $L$,



$$ ⟨_⟩ ,:, (X → H) ,⟶, (L → H) $$



Such that this is indeed an extension when restricted to the elements of $X$,



$$ (0) qquad ϕ ∘ 𝒾 = f quad≡quad ϕ = ⟨f⟩ $$



In (0) taking $ϕ$ to be $⟨f⟩$ yields



$$ (1) qquadqquadqquadqquad ⟨f⟩ ∘ 𝒾 = f $$



In (0) taking $ϕ, f$ to be $id, i$ yields



$$ (2) qquadqquadqquad;;;;qquad id = ⟨𝒾⟩ $$



Now suppose we are given $g : X ≅ Y$ where
$X, Y$ have free groups $L₁, 𝒾₁$ and $L₂, 𝒾₂$, then we
claim $⟨i₂ ∘ g⟩ : L₁ ≅ L₂$ with inverse $⟨𝒾₁ ∘ g⁻¹⟩$.



Indeed we can show this to be true with a simple calculation:
--omitting the subscripts--



 ⟨𝒾 ∘ g⟩ ∘ ⟨𝒾 ∘ g⁻¹⟩ = id
≡ Using (2) with an aim to utilising (0) 
 ⟨𝒾 ∘ g⟩ ∘ ⟨𝒾 ∘ g⁻¹⟩ = ⟨𝒾⟩
≡ Now in a position to utilise (0) 
 ⟨𝒾 ∘ g⟩ ∘ ⟨𝒾 ∘ g⁻¹⟩ ∘ 𝒾 = 𝒾
≡⟨ The only thing we can use is (1), so let's try it }
 ⟨𝒾 ∘ g⟩ ∘ 𝒾 ∘ g⁻¹ = 𝒾
≡⟨ Again the only thing we can use is (1), so let's try it }
 𝒾 ∘ g ∘ g⁻¹ = 𝒾
≡ Inverses and identity laws 
 𝒾 = 𝒾
≡ Reflexivity of equality 
 true





share|cite|improve this answer





















  • Thanks for your approach :)
    – Javi
    Aug 1 at 19:30














up vote
1
down vote













While there is already an accepted answer, I'd like to showcase a more algebraic solution to the problem which is not as common.



That $L, 𝒾$ is the free group for $X$ means that there is a way to
“extend” maps from $X$ to homomorphisms from $L$,



$$ ⟨_⟩ ,:, (X → H) ,⟶, (L → H) $$



Such that this is indeed an extension when restricted to the elements of $X$,



$$ (0) qquad ϕ ∘ 𝒾 = f quad≡quad ϕ = ⟨f⟩ $$



In (0) taking $ϕ$ to be $⟨f⟩$ yields



$$ (1) qquadqquadqquadqquad ⟨f⟩ ∘ 𝒾 = f $$



In (0) taking $ϕ, f$ to be $id, i$ yields



$$ (2) qquadqquadqquad;;;;qquad id = ⟨𝒾⟩ $$



Now suppose we are given $g : X ≅ Y$ where
$X, Y$ have free groups $L₁, 𝒾₁$ and $L₂, 𝒾₂$, then we
claim $⟨i₂ ∘ g⟩ : L₁ ≅ L₂$ with inverse $⟨𝒾₁ ∘ g⁻¹⟩$.



Indeed we can show this to be true with a simple calculation:
--omitting the subscripts--



 ⟨𝒾 ∘ g⟩ ∘ ⟨𝒾 ∘ g⁻¹⟩ = id
≡ Using (2) with an aim to utilising (0) 
 ⟨𝒾 ∘ g⟩ ∘ ⟨𝒾 ∘ g⁻¹⟩ = ⟨𝒾⟩
≡ Now in a position to utilise (0) 
 ⟨𝒾 ∘ g⟩ ∘ ⟨𝒾 ∘ g⁻¹⟩ ∘ 𝒾 = 𝒾
≡⟨ The only thing we can use is (1), so let's try it }
 ⟨𝒾 ∘ g⟩ ∘ 𝒾 ∘ g⁻¹ = 𝒾
≡⟨ Again the only thing we can use is (1), so let's try it }
 𝒾 ∘ g ∘ g⁻¹ = 𝒾
≡ Inverses and identity laws 
 𝒾 = 𝒾
≡ Reflexivity of equality 
 true





share|cite|improve this answer





















  • Thanks for your approach :)
    – Javi
    Aug 1 at 19:30












up vote
1
down vote










up vote
1
down vote









While there is already an accepted answer, I'd like to showcase a more algebraic solution to the problem which is not as common.



That $L, 𝒾$ is the free group for $X$ means that there is a way to
“extend” maps from $X$ to homomorphisms from $L$,



$$ ⟨_⟩ ,:, (X → H) ,⟶, (L → H) $$



Such that this is indeed an extension when restricted to the elements of $X$,



$$ (0) qquad ϕ ∘ 𝒾 = f quad≡quad ϕ = ⟨f⟩ $$



In (0) taking $ϕ$ to be $⟨f⟩$ yields



$$ (1) qquadqquadqquadqquad ⟨f⟩ ∘ 𝒾 = f $$



In (0) taking $ϕ, f$ to be $id, i$ yields



$$ (2) qquadqquadqquad;;;;qquad id = ⟨𝒾⟩ $$



Now suppose we are given $g : X ≅ Y$ where
$X, Y$ have free groups $L₁, 𝒾₁$ and $L₂, 𝒾₂$, then we
claim $⟨i₂ ∘ g⟩ : L₁ ≅ L₂$ with inverse $⟨𝒾₁ ∘ g⁻¹⟩$.



Indeed we can show this to be true with a simple calculation:
--omitting the subscripts--



 ⟨𝒾 ∘ g⟩ ∘ ⟨𝒾 ∘ g⁻¹⟩ = id
≡ Using (2) with an aim to utilising (0) 
 ⟨𝒾 ∘ g⟩ ∘ ⟨𝒾 ∘ g⁻¹⟩ = ⟨𝒾⟩
≡ Now in a position to utilise (0) 
 ⟨𝒾 ∘ g⟩ ∘ ⟨𝒾 ∘ g⁻¹⟩ ∘ 𝒾 = 𝒾
≡⟨ The only thing we can use is (1), so let's try it }
 ⟨𝒾 ∘ g⟩ ∘ 𝒾 ∘ g⁻¹ = 𝒾
≡⟨ Again the only thing we can use is (1), so let's try it }
 𝒾 ∘ g ∘ g⁻¹ = 𝒾
≡ Inverses and identity laws 
 𝒾 = 𝒾
≡ Reflexivity of equality 
 true





share|cite|improve this answer













While there is already an accepted answer, I'd like to showcase a more algebraic solution to the problem which is not as common.



That $L, 𝒾$ is the free group for $X$ means that there is a way to
“extend” maps from $X$ to homomorphisms from $L$,



$$ ⟨_⟩ ,:, (X → H) ,⟶, (L → H) $$



Such that this is indeed an extension when restricted to the elements of $X$,



$$ (0) qquad ϕ ∘ 𝒾 = f quad≡quad ϕ = ⟨f⟩ $$



In (0) taking $ϕ$ to be $⟨f⟩$ yields



$$ (1) qquadqquadqquadqquad ⟨f⟩ ∘ 𝒾 = f $$



In (0) taking $ϕ, f$ to be $id, i$ yields



$$ (2) qquadqquadqquad;;;;qquad id = ⟨𝒾⟩ $$



Now suppose we are given $g : X ≅ Y$ where
$X, Y$ have free groups $L₁, 𝒾₁$ and $L₂, 𝒾₂$, then we
claim $⟨i₂ ∘ g⟩ : L₁ ≅ L₂$ with inverse $⟨𝒾₁ ∘ g⁻¹⟩$.



Indeed we can show this to be true with a simple calculation:
--omitting the subscripts--



 ⟨𝒾 ∘ g⟩ ∘ ⟨𝒾 ∘ g⁻¹⟩ = id
≡ Using (2) with an aim to utilising (0) 
 ⟨𝒾 ∘ g⟩ ∘ ⟨𝒾 ∘ g⁻¹⟩ = ⟨𝒾⟩
≡ Now in a position to utilise (0) 
 ⟨𝒾 ∘ g⟩ ∘ ⟨𝒾 ∘ g⁻¹⟩ ∘ 𝒾 = 𝒾
≡⟨ The only thing we can use is (1), so let's try it }
 ⟨𝒾 ∘ g⟩ ∘ 𝒾 ∘ g⁻¹ = 𝒾
≡⟨ Again the only thing we can use is (1), so let's try it }
 𝒾 ∘ g ∘ g⁻¹ = 𝒾
≡ Inverses and identity laws 
 𝒾 = 𝒾
≡ Reflexivity of equality 
 true






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answered Aug 1 at 15:17









Musa Al-hassy

1,0381710




1,0381710











  • Thanks for your approach :)
    – Javi
    Aug 1 at 19:30
















  • Thanks for your approach :)
    – Javi
    Aug 1 at 19:30















Thanks for your approach :)
– Javi
Aug 1 at 19:30




Thanks for your approach :)
– Javi
Aug 1 at 19:30












 

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