Indeed, Theorem 1B.9 is a good tool here. In particular, you'll want to use the uniqueness part of the theorem. That means that any two maps $(X,x_0)\to (K(G,1),y_0)$ which induce the same map on $\pi_1$ are homotopic. What does that tell you about a map $f:(X,x_0)\to (K(G,1),y_0)$ which induces the trivial homomorphism on $\pi_1$?
The answer is hidden below.
It tells you $f$ is homotopic to the constant map $(X,x_0)\to (K(G,1),y_0)$, since the constant map also induces the trivial homomorphism on $\pi_1$. So, if there are no nontrivial homomorphisms $\pi_1(X)\to G$, this applies to every map $f:X\to Y$ (for an appropriate choice of basepoints $x_0$ and $y_0$).
I believe I have figured out the answer.
To give a "concrete description" of all the $n$-sheeted covering space up to isomorphism, we need to know all the $m$-sheeted covering spaces of $X$ and $Y$ for $m\leq n$. This includes $X$ and $Y$ themselves. Specifically, an $n$-sheeted covering space of $X\vee Y$ is a connected "graph" with the following properties:
- There are $n$ edges
- Given a vertex of our graph of valence $m$, that vertex is an $m$-sheeted connected covering space of either $X$ or $Y$.
- No edge connects two covering spaces of $X$ or two covering spaces of $Y$.
Specifically, at each vertex, the edges attach at points belonging to the preimage of the basepoint $x_0\in X\vee Y$. We can pick any of these edges to be a basepoint of the covering space. Namely, once this graph is constructed, retract each of the lines connecting the vertices that make up the "edges" to points. For example, a graph with two vertices would become the wedge sum of two spaces.
A concrete example: consider the space $B=\mathbb RP^2\vee T^2$ (where $T^2=S^1\times S^1$ is the torus). $B$ has $7$ isomorphism classes of connected two-sheeted covering spaces. This can be seen from the fact that:
- $\mathbb RP^2$ has one two-sheeted covering space, namely, $S^2$.
- $T^2$ has three two-sheeted covering spaces, specifically, those corresponding to the subgroups $\langle a^2,b\rangle$, $\langle a,b^2\rangle$, and $\langle a^2,ba^{-1}\rangle$ of $\pi_1(T^2)=\mathbb Z^2=\langle a,b\mid [a,b]\rangle$ (these can be seen as the kernels of epimorphisms $\mathbb Z^2\to\mathbb Z_2$). Let's call these covering spaces $\widetilde T_1$, $\widetilde T_2$, and $\widetilde T_3$,
There are two possible connected graphs with $2$ edges, namely, that with three vertices obtained by connecting two line-segments end-to-end, and that with two vertices, each vertex of valence 2 with both edges leading to the other vertex.
In the former case, if middle vertex (of valence 2) is a two-sheeted covering space of $\mathbb RP^2$, of which there is only one choice, then the outermost vertices are both $T^2$. On the other hand, there are three possibilities for the middle vertex if it is a covering space of $T^2$, namely, $\widetilde T_1$, $\widetilde T_2$, and $\widetilde T_3$. The outermost vertices must be $\mathbb RP^2$. Hence, there are $4$ possible $2$-sheeted covering spaces of $B$ with three "vertices."
In the latter case, we have two vertices each of valence two. For the vertex that covers $T^2$, we have three choices, and for the vertex that covers $\mathbb RP^2$. we have only one choice, the sphere. In this way, we obtain $3$ more covering spaces of $B$.
I have drawn each of these graphs and labelled the vertices accordingly below. Red vertices correspond to covering spaces of the torus, while blue vertices correspond to covering spaces of $\mathbb RP^2$.
Best Answer
basically it is true for any space $Y$ with fundamental group $G$ s.t there is no non-trivial homomorphism from $\pi_1(X) \to G$...and universal cover of $Y$ is contractible (basically $Y$ is a $K(G,1)$ space)...then by using map lifting criterion you can always get a lifting of your function $f$ on the universal cover where the image is contractible and composition with covering map will give you a null homotopy map for $f$ in $Y$.