# 0708-1300/Class notes for Thursday, January 24

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

INCOMPLETE AND UNEDITED: Completion coming soon.

Announcements go here

## 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(Y_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:

$\begin{matrix} &\ \ \ \ U_1&&\\ &\nearrow^{i_1}&\searrow^{j_1}&\\ U_1\cap U_2&&&U_1\cup U_2 = X\\ &\searrow_{i_2}&\nearrow^{j_2}&\\ &\ \ \ \ U_2&&\\ \end{matrix}$

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.