Precomposition with a faithful functor

Question

If $F: C \rightarrow D$ is a faithful functor, then is the precomposition with $F$ functor $F^{\star}:[D:\mathbf{Set}] \rightarrow [C:\mathbf{Set}]$ faithful?

Answer 1

No. Let $\mathcal{C} = \emptyset$. Then any functor $\mathcal{C} \to \mathcal{D}$ is faithful, while $[\mathcal{D}, \mathbf{Set}] \to [\mathcal{C}, \mathbf{Set}]$ is faithful if and only if $\mathcal{D} = \emptyset$ as well.

The correct condition to ask for is (essential) surjectivity on objects.

Answer 2

Well, let $C=\mathbf{1}$ and $D=\mathbf{2}$, the discrete categories, and $$F:\mathbf{1}→\mathbf{2},\ 1\mapsto 2$$ Then $F$ is vacuously faithful. Define two functor $\mathbf{2}→\mathbf{Set}$ $$G(1)=\{a,b\}=G(2),\qquad H(1)=\{a,b\},\ H(2)=\{a\}$$ Since there are no morphisms in the categories, all collections of maps $\{a,b\}→\{a,b\}, \{a,b\}→\{a\}$ are natural. In particular choose $$σ_1=\text{id}_{\{a,b\}},\ σ_2=a,\ \tau_1=a,\ \tau_2=a$$ Then $$(F^*σ)_1=σ_{F1}=σ_2=\tau_2=(F^*\tau)_1$$ is a single transformation between $GF$ and $G'F$.