If $A$ is a deformation retract of $X$ and $B$ is a deformation retract of $A$ then $ B$ is a deforemation retract of X

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If $A$ is a deformation retract of $X$ and $B$ is a deformation retract of $A$ then $ B$ is a deformation retract of $X$.



I am a beginner in Algebraic Topology so I tried to write every proof out myself before consulting. I know there are various ways of handling this but I just want to confirm if what I wrote below makes sense.



$Asubset X$ is a deformation retract of X, let $r:X to A$ such that $r$ is homotopic to the identity map on $X$.
Define $H:Xtimes I to X$ such that
$$H(x,0)=x$$
$$H(x,1)in A$$
$$H(a,t)=a, ain A$$
Thus $H(x,0)=x$ and $H(x,1)=r(x)$



Similarly, We have $Bsubset A$, a deformation retract of A, Let $s:A to B$ such that $s$ is homotopic to the identity map on $A$.
Define $F:Atimes I to A $ such that
$$F(x,0)=x$$
$$F(x,1)in B$$
$$F(b,t)=B, bin B$$
Thus $F(x,0)=x$ and $F(x,1)=s(x)$



Now to show that $B$ is a deformation retract of $X$, I define
$q=scirc r :X to B$.



so $q$ is continuous being the composition of two continuous functions.



Claim: $G:X times I to X$. define by
$$G(x,t)=F(H(x,t),t)$$ Is the required homotopy between $q$ and the identity on $X$.
$$G(x,0)=F(H(X,0),0)=F(x,0)=x$$
$$G(x,1)=F(H(x,1),1)=F(r(x),1)=s(r(x))=scirc r$$
Is this a good way to go?
Any help will be appreciated. thank you.




algebraic-topology fundamental-groups




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  • This is a good way to go.
    – Tyrone
    Jul 3 at 9:33










  • @Tyrone Thank you.
    – J. Kyei
    Jul 4 at 1:34














up vote
1
down vote

favorite












If $A$ is a deformation retract of $X$ and $B$ is a deformation retract of $A$ then $ B$ is a deformation retract of $X$.



I am a beginner in Algebraic Topology so I tried to write every proof out myself before consulting. I know there are various ways of handling this but I just want to confirm if what I wrote below makes sense.



$Asubset X$ is a deformation retract of X, let $r:X to A$ such that $r$ is homotopic to the identity map on $X$.
Define $H:Xtimes I to X$ such that
$$H(x,0)=x$$
$$H(x,1)in A$$
$$H(a,t)=a, ain A$$
Thus $H(x,0)=x$ and $H(x,1)=r(x)$



Similarly, We have $Bsubset A$, a deformation retract of A, Let $s:A to B$ such that $s$ is homotopic to the identity map on $A$.
Define $F:Atimes I to A $ such that
$$F(x,0)=x$$
$$F(x,1)in B$$
$$F(b,t)=B, bin B$$
Thus $F(x,0)=x$ and $F(x,1)=s(x)$



Now to show that $B$ is a deformation retract of $X$, I define
$q=scirc r :X to B$.



so $q$ is continuous being the composition of two continuous functions.



Claim: $G:X times I to X$. define by
$$G(x,t)=F(H(x,t),t)$$ Is the required homotopy between $q$ and the identity on $X$.
$$G(x,0)=F(H(X,0),0)=F(x,0)=x$$
$$G(x,1)=F(H(x,1),1)=F(r(x),1)=s(r(x))=scirc r$$
Is this a good way to go?
Any help will be appreciated. thank you.




algebraic-topology fundamental-groups




share|cite|improve this question





















  • This is a good way to go.
    – Tyrone
    Jul 3 at 9:33










  • @Tyrone Thank you.
    – J. Kyei
    Jul 4 at 1:34












up vote
1
down vote

favorite









up vote
1
down vote

favorite











If $A$ is a deformation retract of $X$ and $B$ is a deformation retract of $A$ then $ B$ is a deformation retract of $X$.



I am a beginner in Algebraic Topology so I tried to write every proof out myself before consulting. I know there are various ways of handling this but I just want to confirm if what I wrote below makes sense.



$Asubset X$ is a deformation retract of X, let $r:X to A$ such that $r$ is homotopic to the identity map on $X$.
Define $H:Xtimes I to X$ such that
$$H(x,0)=x$$
$$H(x,1)in A$$
$$H(a,t)=a, ain A$$
Thus $H(x,0)=x$ and $H(x,1)=r(x)$



Similarly, We have $Bsubset A$, a deformation retract of A, Let $s:A to B$ such that $s$ is homotopic to the identity map on $A$.
Define $F:Atimes I to A $ such that
$$F(x,0)=x$$
$$F(x,1)in B$$
$$F(b,t)=B, bin B$$
Thus $F(x,0)=x$ and $F(x,1)=s(x)$



Now to show that $B$ is a deformation retract of $X$, I define
$q=scirc r :X to B$.



so $q$ is continuous being the composition of two continuous functions.



Claim: $G:X times I to X$. define by
$$G(x,t)=F(H(x,t),t)$$ Is the required homotopy between $q$ and the identity on $X$.
$$G(x,0)=F(H(X,0),0)=F(x,0)=x$$
$$G(x,1)=F(H(x,1),1)=F(r(x),1)=s(r(x))=scirc r$$
Is this a good way to go?
Any help will be appreciated. thank you.




algebraic-topology fundamental-groups




share|cite|improve this question













If $A$ is a deformation retract of $X$ and $B$ is a deformation retract of $A$ then $ B$ is a deformation retract of $X$.



I am a beginner in Algebraic Topology so I tried to write every proof out myself before consulting. I know there are various ways of handling this but I just want to confirm if what I wrote below makes sense.



$Asubset X$ is a deformation retract of X, let $r:X to A$ such that $r$ is homotopic to the identity map on $X$.
Define $H:Xtimes I to X$ such that
$$H(x,0)=x$$
$$H(x,1)in A$$
$$H(a,t)=a, ain A$$
Thus $H(x,0)=x$ and $H(x,1)=r(x)$



Similarly, We have $Bsubset A$, a deformation retract of A, Let $s:A to B$ such that $s$ is homotopic to the identity map on $A$.
Define $F:Atimes I to A $ such that
$$F(x,0)=x$$
$$F(x,1)in B$$
$$F(b,t)=B, bin B$$
Thus $F(x,0)=x$ and $F(x,1)=s(x)$



Now to show that $B$ is a deformation retract of $X$, I define
$q=scirc r :X to B$.



so $q$ is continuous being the composition of two continuous functions.



Claim: $G:X times I to X$. define by
$$G(x,t)=F(H(x,t),t)$$ Is the required homotopy between $q$ and the identity on $X$.
$$G(x,0)=F(H(X,0),0)=F(x,0)=x$$
$$G(x,1)=F(H(x,1),1)=F(r(x),1)=s(r(x))=scirc r$$
Is this a good way to go?
Any help will be appreciated. thank you.





algebraic-topology fundamental-groups





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Aug 14 at 3:19
























asked Jul 2 at 19:16









J. Kyei

1758




1758











  • This is a good way to go.
    – Tyrone
    Jul 3 at 9:33










  • @Tyrone Thank you.
    – J. Kyei
    Jul 4 at 1:34
















  • This is a good way to go.
    – Tyrone
    Jul 3 at 9:33










  • @Tyrone Thank you.
    – J. Kyei
    Jul 4 at 1:34















This is a good way to go.
– Tyrone
Jul 3 at 9:33




This is a good way to go.
– Tyrone
Jul 3 at 9:33












@Tyrone Thank you.
– J. Kyei
Jul 4 at 1:34




@Tyrone Thank you.
– J. Kyei
Jul 4 at 1:34















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