Surjectivity of Induced Homomorphism in Algebraic Topology

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Discussion Overview

The discussion revolves around the surjectivity of the induced homomorphism r* from the first homotopy group Pi_1(X,a_0) to Pi_1(A,a_0) in the context of algebraic topology. Participants explore the implications of a continuous map r from a space X to a subset A, particularly focusing on the conditions under which r* is surjective.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant expresses uncertainty about how to formally demonstrate the surjectivity of r* and contemplates the need for a path homotopy from a loop in A to a loop in X.
  • Another participant suggests that the loop [f] in A can be used directly as g in X, arguing that since r fixes A, it should map [f] onto itself.
  • There is a reiteration of the idea that f is a loop in X because it is a loop in A, given that A is a subset of X.
  • A participant questions the role of the properties of r in the argument for surjectivity, hinting at a potential connection to a pasting lemma.
  • Some participants express a feeling that the argument for surjectivity may be overly simplistic, while others affirm the reasoning presented.

Areas of Agreement / Disagreement

Participants generally agree that f being a loop in A implies it is also a loop in X. However, there is uncertainty regarding the simplicity of the argument and the role of the properties of r, indicating that the discussion remains unresolved.

Contextual Notes

Participants have not fully resolved the implications of the properties of the map r and how they affect the surjectivity of the induced homomorphism. There are also unresolved questions about the necessity of constructing a path homotopy.

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.
 
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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.
 
DeadWolfe said:
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?
 
f is a loop in X because it is a loop in A, and A is a subset of X.
 
DeadWolfe said:
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?
 
DeadWolfe said:
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.
 
Well, if r did not fix a then we would not have r(f)=f.
 
DeadWolfe said:
Well, if r did not fix a then we would not have r(f)=f.

That's true.
 

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