Induced maps in homology are injective

In a topological course I have learned that a covering map $p:\bar{X}\rightarrow X$ induces a map on fundamental groups and that this induced map is a monomorphism.

Now I wonder myself if this is also true for $H_1(p)$. Via the Hurewicz-isomorphism $H_1(p)$ is isomorphic to the abelianization of the fundamental group.

Solutions Collecting From Web of "Induced maps in homology are injective"

No, this does not need to be the case. Let $X$ be any space whose fundamental group is perfect (i.e. such that its abelianization is zero), for example the classifying space $X = BA_5$ of the alternating group $A_5$. Let $H \subset A_5$ be the subgroup generated by a transposition, for example, $H = \{ 1, \sigma \}$ with $\sigma = (1 \; 2)$. The group $H$ is isomorphic to $\mathbb{Z}/2\mathbb{Z}$, whose abelianization is itself.

Then by the classification of covering spaces, there exists a covering space $p : \tilde{X} \to X$ such that $\pi_1(\tilde{X}) = H$ and $p$ induces the inclusion $H \subset A_5$ on fundamental groups. Now on the level of homology, $H_1(\tilde{X}) = H_{\mathrm{ab}} = H$, whereas $H_1(X) = (A_5)_{\mathrm{ab}} = 0$. Thus the induced map $H_1(\tilde{X}) \to H_1(X)$ cannot be a monomorphism.


Here is another example that doesn’t use classifying spaces. Let $X = S^1 \vee S^1$ be the figure-eight space, the wedge sum of two circles. Its fundamental group is the free group on two generators,
$$\pi_1(X) = \mathbb{Z} * \mathbb{Z} = \langle a,b \rangle.$$

Consider the subgroup $H$ generated by $g = aba^{-1}b^{-1}$. Since $g$ has infinite order, you get $H \cong \mathbb{Z}$.

Using the classification of covering spaces, there exists a covering space $p : \tilde{X} \to X$ such that $\pi_1(\tilde{X}) = H = \mathbb{Z}$ and $p$ induces the inclusion $H \subset \pi_1(X)$ on fundamental groups.

Now, $H$ is abelian, so $H_1(\tilde{X}) = H = \mathbb{Z}$, with generator $[g]$. On the other hand, $H_1(X) = \mathbb{Z} \oplus \mathbb{Z}$ is the direct sum of two copies of $\mathbb{Z}$, with generators $[a]$ and $[b]$. The morphism $p$ induces on $H_1$:
$$p_*([g]) = [a] + [b] – [a] – [b] = 0$$
and so is the zero morphism on homology, which is not injective.