Algebraic Topology - Inclusion into RP^2

[kreewser]

Homework Statement

Find an inclusion map i from S^1 to RP^2 such that the induced map of the inclusion (by the fundamental group) is not the zero element.

Known:
pi_1(S^1) = Z and pi_1(RP^2) = Z/2Z

Homework Equations

Can we define i as a composite of two other inclusions?

The Attempt at a Solution

I was thinking of using that fact that we can regard RP^2= D^2 V Mobius (i.e. as a Wedge Sum over the outer the edge circle of the Mobius band) and try to use a commutative diagram with an inclusion map j from S^1 to Mobius, then an inclusion k from the Mobius band to RP^2 and take the composite i=j(k). Is this valid? (For example, for the induced map j*, can there be j*:Z to Z/2Z ?) .This may be trivial, I don't know...

(I am sorry in advance if I have posted this in the wrong forum page)
Thanks

 

[mighty2000]

for your question and for your attempt at a solution. You are on the right track with your approach. Let's break it down step by step.

First, as you mentioned, we can regard RP^2 as a wedge sum of D^2 and the Mobius band, which we can denote as RP^2 = D^2 V Mobius. This means that we can define an inclusion map k from the Mobius band to RP^2.

Next, we can define an inclusion map j from S^1 to the Mobius band. This is because S^1 is the boundary of the disk D^2, and the boundary of the Mobius band is also S^1.

Now, we can define the composite inclusion map i = j(k) from S^1 to RP^2. This means that i is the inclusion of S^1 into the Mobius band, followed by the inclusion of the Mobius band into RP^2.

To show that this induced map is not the zero element, we need to show that the induced map i* from pi_1(S^1) to pi_1(RP^2) is not the trivial homomorphism. This can be done by showing that there exists a nontrivial element in pi_1(S^1) that maps to a nontrivial element in pi_1(RP^2).

We know that pi_1(S^1) = Z, and pi_1(RP^2) = Z/2Z. This means that the induced map i* can be represented as i*: Z to Z/2Z. We can choose a generator of Z, say 1, and show that i*(1) is a nontrivial element in Z/2Z. This can be done by noting that i*(1) is the image of a nontrivial loop in S^1, which is the boundary of the disk D^2. This nontrivial loop can be chosen to be the boundary of the disk D^2 itself, which maps to the nontrivial loop in RP^2.

Therefore, the induced map i* is not the trivial homomorphism, and thus i is the inclusion map from S^1 to RP^2 with the desired property.

I hope this helps clarify your approach and shows that it is indeed valid. Good luck with your studies!