= Solution
A <group homomorphism> from the additive <cyclic group> $\mathbb Z_6$ is determined by the image $t$ of its generator $1$, with the consistency condition $6t=0$ in the target. Every element of either target satisfies this condition. Therefore
$$
\boxed{\mathbb Z_6\to\mathbb Z_2:\quad f_t([n]_6)=[tn]_2,\quad t=0,1,}
$$
and
$$
\boxed{\mathbb Z_6\to\mathbb Z_3:\quad f_t([n]_6)=[tn]_3,\quad t=0,1,2.}
$$
These formulas are well-defined because changing $n$ by six changes $tn$ by a target multiple of its modulus. They list respectively two and three maps.
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