Algebraic Topology Questions

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  • #1
JasonRox
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I'm totally stuck on these two.

The first is...

Let A be a subset of X; suppose r:X->A is a continuous map from X to A such that r(a)=a for each a e A. If a_0 e A, show that...

r* : Pi_1(X,a_0) -> Pi_1(A,a_0)

...is surjective.

Note: Pi_1 is the first homotopy group and r* is the homomorphism induced by h.

I can visually see in my mind why this is so, but I can't even think of how to write this down at all.

I'm still thinking about it. No need to post anything right now.

The way I'm thinking that is if [f] is in A then I need to show that there is a g in X such that g is path homotopic to f using probably r to create my path homotopy. (f is a loop around a_0 in A)

Once I do that, then it should come out like... r([g]) = [r o g] = [f].

I'm still thinking about this.
 

Answers and Replies

  • #2
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Can't you use [f] itself for g? I mean, if [f] is a loop in A, then it is a loop in X, and since r fixes A, it should map [f] onto itself.
 
  • #3
JasonRox
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Can't you use [f] itself for g? I mean, if [f] is a loop in A, then it is a loop in X, and since r fixes A, it should map [f] onto itself.
That's exactly what I was thinking too!

But you must show that f is a loop in X by creating the path homotopy that I was speaking about from f to g. Isn't that right?
 
  • #4
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f is a loop in X because it is a loop in A, and A is a subset of X.
 
  • #5
JasonRox
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f is a loop in X because it is a loop in A, and A is a subset of X.
So, to show it is surjective, we have...

r([f]) = [r o f] = [f]

If that is so, where does the properties of r even play a role in here?

I feel like there is a pasting lemma in here or something.

Is it really that simple?
 
  • #6
JasonRox
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f is a loop in X because it is a loop in A, and A is a subset of X.
I do understand this.

For some reason, I feel like it's a little too simple.
 
  • #7
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Well, if r did not fix a then we would not have r(f)=f.
 
  • #8
JasonRox
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Well, if r did not fix a then we would not have r(f)=f.
That's true.
 

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