0708-1300/Class notes for Thursday, January 24

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Typed Notes

The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.

Proof of Van Kampen

Let G_i = \pi_1(U_i)

H = \pi_1(U_1\cap U_2)

G=\pi_1(U_1\cup U_2)

We aim to show that G=G_1*_H G_2

Hence, we want to define two maps:

\Phi:G_1*_H G_2\rightarrow G and

\Psi:G\rightarrow G_1*_H G_2

such that they are inverses of each other.

Now, recall the commuting diagram:

&\ \ \ \  U_1&&\\
U_1\cap U_2&&&U_1\cup U_2 = X\\
&\ \ \ \  U_2&&\\

Further, let b_i alternate between 1 and 2 for successive i's.

Hence we define \Phi via for \alpha_i\in G_{b_i},

[\alpha_1][\alpha_2]\ldots[\alpha_n]\rightarrow [j_{b_1 *}\alpha_1\cdot\ldots\cdot j_{b_n *}\alpha_n]

Clearly this is well defined. We need to check the relations in G_1*_H G_2 indeed hold. Well, the identity element corresponds to the identity path so the relation that removes identities holds. Furthermore, the concatenation of paths is a sum after \Phi and the definition necessitates the third relation holds.

Now for \Psi:

Elements in G correspond with paths \gamma in U_1\cup U_2

On consider such a \gamma. The Lesbegue Lemma let us break \gamma up so \gamma = \gamma_1\ldots\gamma_N such that each \gamma_i is in just G_{b_i}

Now, \gamma_i does not go from base point to base point, so we can't consider it as a loop itself. Let x_i denote the endpoint of \gamma_i (and hence x_{i-1} is the beginning point) and further, x_0 = b the base point.

Choose paths \eta_i connecting x_i to b such that if x_i\in U_{b_i} then \eta_i\sub U_{b_i}

Hence, \gamma\sim\bar{\eta_0}\gamma_1\eta_1\bar{\eta_1}\gamma_2\eta_2\ldots \mapsto [\bar{\eta_0}\gamma_1\eta_1][\bar{\eta_1}\gamma_2\eta_2]\ldots

where each section is entirely in G_{b_i}. This above mapping is \Psi

We need to show that \Psi is well defined. I.e.,

1) \Psi is independent of the subdivision

2) \Psi is independent of choice of the \eta_i's

3) If \gamma_1\sim\gamma_2 then \Psi(\gamma_1) = \Psi(\gamma_2)

For 1) it is enough to show that one can add or remove a single subdivision point. Suppose you add a new subdivision point c in between two others a and b. Traveling between a and b is the same as going from a to c then to the basepoint and back and then continuing on to b. But this is precisely what happens when you add a subdivision point. Likewise for removing points.

For 2), Suppose \eta_i goes from the basepoint to x_i. Add a subdivision point y right beside x_i and remove the one at x_i. Then add again the original point x_i. The trick is that the new \eta_i' is the one used when adding y. The map from the new x_i to the basepoint is this \eta_i' with the infinitesimal connection between y and x_i added. The subdivision point y is then removed leaving our original configuration of basepoints, only with the path from x_i to b now being \eta_i' instead of the original \eta_i

For 3) we consider the homotopy between \gamma_1 and \gamma_2 thought of as a square with \gamma_1 on the bottom and \gamma_2 on the top. We use the Lesbegue Lemma to subdivide the square into many subsquares such that each one is entirely in U_{b_i}

We further modify the homotopy H to a new homotopy \tilde{H} such that each grid point gets mapped to b.

This can be thought of as "pinching" each gird point and pulling it to b or, alternatively, as tossing a "handkerchief on a bed of nails"

If we let \gamma_1 be broken in to section \alpha_1,\ldots, \alpha_n along the bottom then \tilde{H} first lifts \alpha_1 to the top and right sides of the subsquare such that the gridpoint goes to b. \tilde{H} then moves the next square up in an analogous manner until we are at the top with \gamma_2.