Proving Triviality of pi_1(S^n;e) in Algebraic Topology

[bham10246]

Please read the following problem first:

Suppose n > 1 and let S^n be the n-sphere in R^{n+1}. Let e be the unit-coordinate vector (1,0,...,0) on S^n. Prove that the fundamental group pi_1(S^n;e) is the trivial group.


Okay, now my question is what does the notation "pi_1(S^n;e)" mean?

I understand the fundamental group but I don't understand the "semi-colon and then an element"-part.

Thank you!

 

[Jimmy Snyder]

As pi_1(S^n) is the group of equivalence classes of loops in S^n, pi_1(S^n;e) is the group of equivalence classes of loops in S^n that start and end at the point e.

 

[bham10246]

Thank you so much jimmysnyder for clearing that up for me. I looked through Allen Hatcher's book but it wasn't helpful.

So for our example above, is pi_1(S^n) isomorphic to pi_1(S^n;e)? Do you think I need to show that they are isomorphic or can I assume this as "obvious"?

 

[Jimmy Snyder]

I looked through Allen Hatcher's book but it wasn't helpful.

It's in his book entitled "Algebraic Topology". It's hidden on page 26, just above Proposition 1.3 with no indication that it is a definition.

So for our example above, is pi_1(S^n) isomorphic to pi_1(S^n;e)? Do you think I need to show that they are isomorphic or can I assume this as "obvious"?

I don't know. But if a space X is path connected, then there is no difference between $\pi_1(X)$ and $\pi_1(X; x)$ for all $x \in X$.

 

[bham10246]

Hi jimmysnyder, I think what was confusing (at first) is that the notation on these practice problems used a semi-colon, instead of a comma. Semi-colon can mean Homology with Coefficients, which I didn't know.

Yes, I agree that there is no difference between $\pi_1(X; x)$ and $\pi_1(X; e)$ but I think I need to prove that these two groups are really isomorphic by using basepoint change homomorphism (Propositon 1.5 page 28 of Hatcher), don't you think so? And after doing that, re-prove that $\pi_1(X; x)=0$ (Prop 1.14 page 35)?

This is a lot of work but I'm not sure what we're allowed to assume.

Also, there's another problem that I came across:

Let $I = [0,1]$. Let $X$ be a space, and let $p$ and $q$ be two points of $X$. Give an example of a connected space $X$ and points $p$ and $q$ such that $\pi_1(X;p)$ is not isomorphic to $\pi_1(X;q)$.

So $\pi_1(X)$ does depend on the basepoint!

 

[Jimmy Snyder]

So $\pi_1(X)$ does depend on the basepoint!

I don't have time right now to look at it. There is a difference between connected and path connected. This is the red herring rule in mathematics. A red herring is neither red, nor is it a herring.

 

[Jimmy Snyder]

It is easy to see that for any space X, $\pi_1(X; x) \subset \pi_1(X)$ since every loop that begins and ends at x is a loop. Since by Proposition 1.14 (page 35), $\pi_1(S^n) = 0$ for $n\ge2$, most of the work is done. Note that $S^0$ is not path connected, and $\pi_1(S^0;1) \ne \pi_1(S^0)$. That is why in the statement of the problem, n > 1.

I don't have a formal proof that $\pi_1(X;x) = \pi_1(X) \ \forall x \in X$ when X is path connected. But informally, given a loop in X, there is a loop that starts at x, goes to the start point of the loop, loops back to the start point and then goes back to x (because X is path connected). so $\pi_1(X) \subset \pi_1(X; x)$.

It seems you have no more to do than to prove that $S^1$ is path connected to show that $\pi_1(S^1;1) = \pi_1(S^1) = Z$. So that group is not trivial.

 

[bham10246]

Thank you so much for your help!