0708-1300/Class notes for Thursday, January 24

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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. 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 $\eta _{i}\subset U_{b_{i}}$

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 $\gamma _{2}$.