About covering spaces

Question

Suppose X is a topological space whose fundamental group is Z x Z x Z2 x Z3. Is it possible for the wedge sum of two circles to be a covering space for X? Can anyone help me with this ?

Answer 1

A covering space $Y \to X$ induces an injection of fundamental groups $\pi_1(Y) \hookrightarrow \pi_1(X)$. If $Y = S^1 \vee S^1$, then $\pi_1(Y) = \Bbb Z * \Bbb Z$; this is not abelian, and thus cannot be a subgroup of your $\pi_1(X)$, which is. So no, such a covering map is not possible.

In any case, the wedge of two circles doesn't cover any other spaces. If there were a free $G$-action on $S^1 \vee S^1$, any nontrivial element of $G$ would determine a fixed-point-free homeomorphism $S^1 \vee S^1 \to S^1 \vee S^1$. But the wedge point of $S^1 \vee S^1$ is a cut point (a cut point $x \in X$ is a point such that $X \setminus \{x\}$ is disconnected), while no other points are, and homeomorphisms take cut points to cut points. So the wedge point is fixed by any homeomorphism.