Surjectivity of Induced Homomorphism in Algebraic Topology

[JasonRox]

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.

 

[DeadWolfe]

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.

 

[JasonRox]

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?

 

[DeadWolfe]

f is a loop in X because it is a loop in A, and A is a subset of X.

 

[JasonRox]

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?

 

[JasonRox]

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.

 

[DeadWolfe]

Well, if r did not fix a then we would not have r(f)=f.

 

[JasonRox]

Well, if r did not fix a then we would not have r(f)=f.

That's true.