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MathematicalPhysicist

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1.[tex]\pi_1(X,x_0)[/tex] is the fundemental group of X based on x_0, i.e the groups of homotopy classes which are loops based in x0.

Suppose that x1,x0 are points in a path connected space X, then the above group is abelian iff for every pair of paths: a,b from x0 to x1 we get that a'([f])=[a^-1]*[f]*[a]=[b^-1]*[f]*

**=b'([f]) for every [f] in the above group.**

2. Let A be a subset of X, suppose r:X->A is a continuous map s.t r(a)=a for each a in A. if a0 is in A, show that:

[tex]r_*:\pi_1(X,a_0)->\pi_1(A,a_0)[/tex]

is surjective where: [tex]r_*([f])=[rof][/tex]

3. show that every star convex set is simply connected, i.e it's path connected and has the trivial fundemental group.

here's what I did for 1:

suppose the group is abelian, if I show that a'^-1ob' is the identity then it will follow that a'=b' ( or so me think).

so here it goes:

b'^-1oa'([f])=

2. Let A be a subset of X, suppose r:X->A is a continuous map s.t r(a)=a for each a in A. if a0 is in A, show that:

[tex]r_*:\pi_1(X,a_0)->\pi_1(A,a_0)[/tex]

is surjective where: [tex]r_*([f])=[rof][/tex]

3. show that every star convex set is simply connected, i.e it's path connected and has the trivial fundemental group.

here's what I did for 1:

suppose the group is abelian, if I show that a'^-1ob' is the identity then it will follow that a'=b' ( or so me think).

so here it goes:

b'^-1oa'([f])=

**[a^-1][f][a][b^-1]**

but because the group is abelian and b*a^-1 (im using here the notation of my lecturer and munkres for pasting of paths) is a loop based in x_0, then [b*a^-1] commutes with [f] and it equalsbut because the group is abelian and b*a^-1 (im using here the notation of my lecturer and munkres for pasting of paths) is a loop based in x_0, then [b*a^-1] commutes with [f] and it equals

**[a^-1] so the above term equals: [f]****[a^-1][a][b^-1]=[f]**

so it's indeed the identity function and thus a'=b'.

to prove the next implication we simply notice that for every paths: a,b from x0 to x1,

a'=b' so if [g] is in this group we can represnet it as:

g=a*b^-1

so if we operate this [f] we will get agian [f], i.e

g'([f])=[g^-1][f][g]=so it's indeed the identity function and thus a'=b'.

to prove the next implication we simply notice that for every paths: a,b from x0 to x1,

a'=b' so if [g] is in this group we can represnet it as:

g=a*b^-1

so if we operate this [f] we will get agian [f], i.e

g'([f])=[g^-1][f][g]=

**[a^-1][f][a][b^-1]=b'^-1(a'([f]))=[f]**

so they commute and it's abelian, am i right here?

now for two, if it acts only or pi_1(A,a_0) then obviously it's surjcetive but how to show it for other elements of pi_1(X,a_0), wait a minute I think that it's simple enough because every element of pi_1(A,a_0) is in the form of [rof] for f a loop based on a_0, cause if it's a loop, the we can paste it with another loop in A based in a_0, and thus every loop based in a_0 in pi_1(X,a_0).

Not sure if i ogt it right i mean if [f] is in pi_1(A,a_0) then obviously r_*([f])=[rof]=[f] so in this case it's the identity but if for example [f] is in pi_1(X,a_0) and not in pi_1(A,a_0), then maybe because rof:[0,1]->A, and rof(0)=r(f(0))=r(a_0)=a_0=rof(1)=r(a_0) then also [rof] is homotopy loop class based in a_0, well now that I type this I think that this is rather rudimentary.

Anyway for the third question I don't see a clear answer, I need to show that every loop based in the star point is the trivial one, i.e it goes and returns the same path but with different directions.

Not sure how to show it, I mean can't we have a loop that passes through three distinct points and it passes by the star points twice and the other two only once (which is a non trivial loop).

any hints?so they commute and it's abelian, am i right here?

now for two, if it acts only or pi_1(A,a_0) then obviously it's surjcetive but how to show it for other elements of pi_1(X,a_0), wait a minute I think that it's simple enough because every element of pi_1(A,a_0) is in the form of [rof] for f a loop based on a_0, cause if it's a loop, the we can paste it with another loop in A based in a_0, and thus every loop based in a_0 in pi_1(X,a_0).

Not sure if i ogt it right i mean if [f] is in pi_1(A,a_0) then obviously r_*([f])=[rof]=[f] so in this case it's the identity but if for example [f] is in pi_1(X,a_0) and not in pi_1(A,a_0), then maybe because rof:[0,1]->A, and rof(0)=r(f(0))=r(a_0)=a_0=rof(1)=r(a_0) then also [rof] is homotopy loop class based in a_0, well now that I type this I think that this is rather rudimentary.

Anyway for the third question I don't see a clear answer, I need to show that every loop based in the star point is the trivial one, i.e it goes and returns the same path but with different directions.

Not sure how to show it, I mean can't we have a loop that passes through three distinct points and it passes by the star points twice and the other two only once (which is a non trivial loop).

any hints?