# 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)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:

- 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:

- So there exists unique lifts $\tilde{p \circ \alpha}$ and $\tilde{p \circ \beta}$ starting at $\tilde{x}$ and such that:

- 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.