(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).