# 5.02 Fundamental group of a CW complex

## Video

Below the video you will find accompanying notes and some pre-class questions.

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**Van Kampen's theorem**. - Next video:
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## Notes

### Example of the 2-torus revisited

*(0.00)* In the previous video, we saw how to use Van Kampen's theorem
to compute the fundamental group of the 2-torus. The result was
\[\langle a,b\ |\ b^{-1}a^{-1}ba=1\rangle.\] There is a CW structure
on the 2-torus (square with opposite sides identified) with one 0-cell
(vertex), two 1-cells (edges) and one 2-cell (the square itself). It
is no coincidence that the presentation we obtained has:

- one generator for each 1-cell,
- one relation for each 2-cell: the relation comes from looking at the boundary of the 2-cell.

### Fundamental group of a CW complex

*(1.24)* If \(X\) is a CW complex with one 0-cell (for simplicity)
then \(\pi_1(X)\) has a presentation where the generators are the
1-cells and the relations come from the 2-cells. More precisely,

- each 1-cell is a loop (as both ends attach to the unique 0-cell). These loops generate \(\pi_1(X)\);
*(3.05)*the boundary of each 2-cell \(e\) gives a loop \(\partial e\) in the 1-skeleton which we can write as a word in the 1-cells; since this loop is nullhomotopic in \(e\) we need the relation \(\partial e=1\) in \(\pi_1(X)\). These relations suffice to give a presentation of \(\pi_1(X)\).

*(5.18)* Note that this theorem is about cell complexes of *any*
dimension; in other words, the 3-cells, 4-cells, etc. do not affect
the fundamental group.

*(5.38)* Let \(X^1\) be the 1-skeleton of \(X\). This is obtained by
attaching 1-cells to a point, so \(X^1\) is just a wedge of 1-cells
\(X^1=\bigvee_{1\mbox{-cells}}S^1\). By induction, using the
computation from last time,
\(\pi_1(X^1)=\bigstar_{1\mbox{-cells}}\mathbf{Z}\) (i.e. the free
product of copies of the integers, one for each 1-cell, in other
words, the group with one generator for each 1-cell and no
relations).

*(7.42)* What happens when we attach a 2-cell?

**Claim:** If \(Y\) is any space and \(Z\) is obtained from \(Y\) by
attaching a 2-cell \(e\) with some attaching map \(\varphi\) then
\(\pi_1(Z)=\pi_1(Y)/N(\partial e)\), where \(N(\partial e)\) is the
normal subgroup generated by the boundary of \(e\) (that is
\([\varphi]\in\pi_1(Y)\)). This is equivalent to imposing the
relation \(\partial e=1\).

*(9.48)* **Proof of claim:** This follows from Van Kampen's
theorem. If we decompose \(Z=U\cup V\) where \(U\) is the interior
of the 2-cell \(int(e)\) and \(V\) is the union \(Y\cup
(e\setminus\{0\})\) (that is, \(V\) is everything except one point
in the interior of \(e\)).

- The intersection \(U\cap V\) is \(e\setminus\{0\}\), the punctured 2-cell. That is homotopy-equivalent to a circle.
- \(U\) is contractible.
- \(V\) is homotopy equivalent to \(Y\) (retracting the punctured 2-cell down onto its boundary in \(Y\)).

*(11.45)* Van Kampen's theorem tells us that
\(\pi_1(Z)=\pi_1(Y)\star_{\mathbf{Z}}\{1\}\). In other words, we
take \(\pi_1(Y)\) and we add an amalgamated relation \(\partial
e=1\) (\(\partial e\) generates \(\pi_1(U\cap V)\): in \(\pi_1(U)\)
is becomes trivial and in \(\pi_1(V)\) it becomes \(\partial e\)).

*(13.28)* What about adding higher dimensional cells? By the same
argument, where \(e\) is an \(n\)-cell with \(n>2\), we get
\[\pi_1(Z)=\pi_1(Y)\star_{\{1\}}\{1\},\]
because \(U\cap V\) is now homotopy equivalent to the
\((n-1)\)-sphere, which is simply-connected (so there is no new
amalgamated relation).

## Pre-class questions

Given that the quotient of the octagon by the identifications indicated in the figure below is a genus 2 surface, use Van Kampen's theorem to give a presentation for the fundamental group of a genus 2 surface.

## Navigation

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**Van Kampen's theorem**. - Next video:
**Fundamental group of a mapping torus**. - Index of all lectures.