= Solution
Whenever a combinatorial loop $\alpha$ traverses an oriented edge and immediately traverses the same edge backwards, delete that backtracking pair. Each deletion is a <based homotopy>[homotopy relative to endpoints] inside $Z$ and reduces the edge length by two, so the process terminates at a reduced, hence locally injective, combinatorial loop $\beta$.
The universal cover $\widetilde Y$ of a connected graph is a <tree>. The lift $\widetilde\beta$ is also locally injective because a <covering space>[covering map] is a local graph isomorphism. A locally injective edge path in a tree cannot repeat a vertex: the segment between two successive visits would be a nonempty reduced closed path, whereas every closed path in a tree backtracks. Thus $\widetilde\beta$ is injective unless $\beta$ is constant.
Now let a loop in $Z$ become null-homotopic in $Y$. Its reduced representative $\beta$ lifts to a closed path in $\widetilde Y$. The preceding injectivity forces that lift, and hence $\beta$, to be constant. The original loop is null-homotopic in $Z$, proving that
$$
\pi_1(Z,y_0)\longrightarrow\pi_1(Y,y_0)
$$
is an injective <group homomorphism>.
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