Proposition

If $X_0$ is the path-component of a space $X$ containing the basepoint $x_0$, show that the inclusion $i: X_0 \rightarrow X$ induces an isomorphism $\pi_1(X_0, x_0) \rightarrow \pi_1(X, x_0)$.

Solution

Since the inclusion $i$ is a continuous function, \(i_*\) is a homomorphism. We will show that it is bijective.

• Injective?

Choose $[f] \in \pi_1(X_0, x_0)$ such that \(i_*([f]) = [e_{x_0}]\) where $e_{x_0}$ denotes the constant loop at $x_0$. Since \(i_*([f]) = [f]\), $f$ and $e_{x_0}$ are homotopic in $X$. Let $f_t: I \rightarrow X$ be a homotopy between $f$ and $e_{x_0}$ in $X$. Let $t_0 \times s_0 \in I \times I$ be given. Then the point $f_{t_0}(s_0)$ is path-connected to $f_{0}(s_0)$ by the path $t \mapsto f_{tt_0}(s_0)$. The point $f_0(s_0)$ is path-connected to $f_0(0) = f(0) = x_0$ by the path $t \mapsto f_0(ts_0)$. Therefore, $f_{t_0}(s_0)$ is path-connected to $x_0$, so $f_{t_0}(s_0) \in X_0$.

This means that $f_t$ maps $I$ into $X_0$. Thus $f_t$ is a homotopy in $X_0$, so $f$ is homotopic to the constant loop at $x_0$ in $X_0$. Therefore, \(i_*\) is injective.

• Surjective?

Let $[f] \in \pi_1(X, x_0)$. Then for each $t_0 \in I$, $f(t_0)$ is path-connected to $f(0)$ by the path $t \mapsto f(tt_0)$. Thus $f$ maps $I$ into $X_0$. Let $g: I \rightarrow X_0$ such that $g(t) = f(t)$. Then \(i_*([g]) = [i \circ g] = [f]\). Therefore, \(i_*\) is surjective.

Hence, \(i_*\) is bijective.