Hatcher exercise 1.1.13 Solution

algebraic-topologyhomotopy-theorysolution-verification

Exercise 1.1.13

Given a space $X$ and a path-connected subspace $A$ containing the base point $x_0$, show that the map $\pi_1 (A,x_0) \to \pi_1 (X,x_0)$ induced by the inclusion $A \hookrightarrow X$ is surjective iff every path in $X$ with endpoints in $A$ is homotopic to a path in $A$.

Solution

$(\Rightarrow)$
Assume that $i_* :\pi_1 (A,x_0) \to \pi_1 (X,x_0)$ is surjective. Let $x_0,x_1 \in A$ and $h:I \to X$ a path, from $x_1$ to $x_0$. Because $A$ is path connected, there is a $g:I\to A$ from $x_0$ to $x_1$. Observe that $\tilde{h} \cdot \tilde{g}$ is a loop in $X$ based in $x_0$. So $\left[ \tilde{h} \cdot \tilde{g} \right] \in \pi_1 (X,x_0)$. By surjective of $i_*$, there is a loop $\left[\alpha \right] \in \pi_1 (A,x_0)$, such that $i_*\left[ \alpha \right] = \left[ \tilde{h} \cdot \tilde{g} \right] \Rightarrow i\circ \alpha \simeq \tilde{h} \cdot \tilde{g} \Rightarrow \alpha \cdot g \simeq_p \tilde{h}$. Where, $\alpha \cdot g $ is a path in $A$, from $x_0$ to $x_1$. Here "$\simeq_p$" means patth homotopy, while "$\simeq$" means homotopy of loops.

$(\Leftarrow)$

Observe that, $i_*$ be surjective means that $\forall \left[ \beta \right] \in \pi_1 (X,x_0),\quad \exists \left[ \alpha \right] \in \pi_1 (A,x_0)$ such that $i_*\left[ \alpha \right] = \left[ \beta \right]$, but $i_*\left[ \alpha \right] = \left[ i \circ \alpha \right] = \left[ \alpha \right] = \left[ \beta \right]$. This means that every loop with base point $x_0$ in $X$ is homotopy to some loop in $A$, with base point $x_0$. But, also note that a loop is a path, and by hypothesis every path in $X$ with endpoint in $A$ is homotopic to a path in $A$. So, by the observation above, $i_*$ is surjective.

Best Answer

This is correct. The only place you have to make sure you understand what is going on is the "$i\circ \alpha \simeq \tilde{h} \cdot \tilde{g} \Rightarrow \alpha \cdot g \simeq_p \tilde{h}$" part, where you "whisker" a homotopy of loops with a path ($g$) and then compose it with the homotopy exhibiting that $\tilde{g} g $ is "the same" as a constant path.