Proof:

Let $\gamma_1$ and $\gamma_2$ be two loops in $X$ based at $x_0$. We have to show that the one chain

$$\gamma_1 + \gamma_2 - \gamma_1*\gamma_2$$

is a boundary of some singular two chain $\sigma$. The idea behind the construction is simple. We first define a map ${\tilde F}:I\times I\longrightarrow X$ whose restrictions to the four sides of the square are $\gamma_1$, $\gamma_2$, $\varepsilon_{x_0}$ and $\gamma_1 * \gamma_2$. As in the previous lemma we shall employ the maps $T_1$, $T_2$ to construct our two chain $\sigma$.

We proceed as in lecture 7 by defining ${\tilde F}$ from the boundary of $I \times I$ to $[0, 1]$, using Tietze's theorem to extend it to the whole of $I \times I$ and then composing with $\gamma_1 * \gamma_2$. So we define

$${\tilde F}(0, s) = \frac{s}{2}$$

$${\tilde F}(t, 1) = \frac{t+1}{2}$$

$${\tilde F}(t, 0) = t$$

$${\tilde F}(1, s) = 1$$

and extend it continuously to $I \times I$. Let $F:I\times I\longrightarrow X$ be given by $F = (\gamma_1*\gamma_2)\circ {\tilde F}.$ The figure below depicts $F$ along the boundary of $I^2$:

in It is now an easy matter to verify that the boundary of the two chain $\sigma$ given by

$$\sigma = F\circ T_1 - F\circ T_2$$

is the one chain

$$\gamma_1 + \gamma_2 - \varepsilon_{x_0} - \gamma_1*\gamma_2 \eqno(32.5)$$

Now, if $\sigma^{\prime}:\Delta_2\longrightarrow X$ be the constant map taking value $x_0$ then $\partial \sigma^{\prime} = \varepsilon_{x_0}$ whereby we conclude that

$$\gamma_1 + \gamma_2 - \gamma_1*\gamma_2 = \partial \sigma + \partial \sigma^{\prime}, \eqno(32.6)$$

which implies $\Pi_{X}([\gamma_1][\gamma_2]) = \Pi_X([\gamma_1*\gamma_2]) = \Pi_X([\gamma_1]) + \Pi_X([\gamma_2])$. $\square$