The Fundamental Group of Spaces and Their Covers

The Fundamental Group of Spaces and Their Covers

Recall from The Induced Mapping from the Fundamental Groups of Two Topological Spaces page that if $X$ and $Y$ are topological spaces and $f : X \to Y$ is a continuous function such that $f(x) = y$ then the induced mapping if the map $f_* : \pi_1(X, x) \to \pi_1(Y, f(x))$ defined for all homotopy classes $[\alpha] \in \pi_1(X, x)$ by:

(1)
\begin{align} \quad f_*([\alpha]) = [f \circ \alpha] \end{align}

We proved that $f_*$ is always a group homomorphism.

When $X$ is a topological space and $(\tilde{X}, p)$ is a covering space, and for a fixed $x \in X$ we choose a $\tilde{x} \in p^{-1}(x)$, then we can look at the induced mapping $p_* : \pi_1(\tilde{X}, \tilde{x}) \to \pi_1(X, x)$.

 Theorem 1: Let $X$ be a topological space and let $(\tilde{X}, p)$ be a covering space of $X$. Let $x \in X$ and let $\tilde{x} \in p^{-1}(x)$. Then $p_* : \pi_1 (\tilde{X}, \tilde{x}) \to \pi_1(X, x)$ is an injective homomorphism.
• Proof: We have already proved that $p_*$ is a homomorphism. We now show that $p_*$ is injective under these conditions.
• Let $[\alpha], [\beta] \in \pi_1(\tilde{X}, \tilde{x})$ (where $\alpha$ and $\beta$ are loops at $\tilde{x}$) and suppose that:
(2)
\begin{align} \quad p_*([\alpha]) = p_*([\beta]) \end{align}
• Then $[p \circ \alpha] = [p \circ \beta]$, so $p \circ \alpha$ and $p \circ \beta$ are loops at $x$ and are homotopic relative to $\{ 0, 1 \}$, that is:
(3)
\begin{align} \quad p \circ \alpha \simeq_{\{0, 1\}} p \circ \beta \end{align}
• So there exists unique lifts $\tilde{p \circ \alpha}$ and $\tilde{p \circ \beta}$ starting at $\tilde{x}$ and such that:
(4)
\begin{align} \quad \tilde{p \circ \alpha} \simeq_{\{0, 1\}} \tilde{p \circ \beta} \end{align}
• But by uniqueness, $\alpha$ and $\beta$ must be these lifts. So $\alpha \simeq_{\{0, 1\}} \beta$ and hence $[\alpha] = [\beta]$. Therefore $p_* : \pi_1(\tilde{X}, \tilde{x}) \to \pi_1(X, x)$ is injective. $\blacksquare$

Recall that if $f : G \to H$ is a group homomorphism then $f(G)$ is a subgroup of $H$.

 Definition: Let $X$ be a topological space and let $(\tilde{X}, p)$ be a covering space of $X$. Let $x \in X$ and let $\tilde{x} \in p^{-1}(x)$. Then we said that $p_*(\pi_1(\tilde{X}, \tilde{x}))$ corresponds to a subgroup of $\pi_1(X, x)$.

For example, let $X$ be a topological space and let $(\tilde{X}, p)$ be a cover. If $\tilde{X} = X$ (the trivial cover) then $p_*(\pi_1(\tilde{X}, \tilde{x}))$ corresponds to the whole space $\pi_1(X, x)$.

If $\tilde{X}$ is a universal cover of $X$ (note, a universal cover need not exist, so we will assume for now that $X$ has a universal cover) then $p_*(\pi_1(\tilde{X}, \tilde{x}))$ corresponds to the trivial group.