# Algebraic topology

1. Jul 11, 2007

### bham10246

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!

2. Jul 11, 2007

### 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.

3. Jul 11, 2007

### 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"?

4. Jul 11, 2007

### Jimmy Snyder

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.

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 $x \in X$.

Last edited: Jul 11, 2007
5. Jul 11, 2007

### 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!

Last edited: Jul 11, 2007
6. Jul 12, 2007

### Jimmy Snyder

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.

Last edited: Jul 12, 2007
7. Jul 12, 2007

### 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.

Last edited: Jul 12, 2007
8. Jul 12, 2007

### bham10246

Thank you so much for your help!