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No. Let $C$ be any category with a zero object. If $F : C \to \text{Set}$ is a cosheaf, then in particular it sends the zero object to the empty set. But since the zero object is also the terminal object, every object in $c$ is equipped with a morphism $c \to 0$, from which it follows that $F(c)$ is equipped with a morphism to the empty set. Hence $F(c)$ is itself empty, so $F$ is the trivial cosheaf. Now take $C$ to be, say, the category of finite-dimensional vector spaces.

Edit: In fact the free example suffices. Let $C$ be the free category on a pair of parallel morphisms $f, g : c \to d$. Then the coproduct $d \sqcup d$ exists and both inclusions $d \to d \sqcup d$ are isomorphisms. It follows that if $F$ is a cosheaf then $F(d)$ is the empty set, from which it follows that $F(c)$ must also be the empty set and $F(f) = F(g)$ must be the unique morphism $\emptyset \to \emptyset$, so $F$ is again the trivial cosheaf.

No. Let $C$ be any category with a zero object. If $F : C \to \text{Set}$ is a cosheaf, then in particular it sends the zero object to the empty set. But since the zero object is also the terminal object, every object in $c$ is equipped with a morphism $c \to 0$, from which it follows that $F(c)$ is equipped with a morphism to the empty set. Hence $F(c)$ is itself empty, so $F$ is the trivial cosheaf. Now take $C$ to be, say, the category of finite-dimensional vector spaces.

Edit: In fact the free example suffices. Let $C$ be the free category on a pair of parallel morphisms $f, g : c \to d$. Then the coproduct $d \sqcup d$ exists and both inclusions $d \to d \sqcup d$ are isomorphisms. It follows that if $F$ is a cosheaf then $F(d)$ is the empty set, from which it follows that $F(c)$ must also be the empty set and $F(f) = F(g)$ must be the unique morphism $\emptyset \to \emptyset$, so $F$ is again the trivial cosheaf.

Edit #2: Here's another example. Let $C = \text{CRing}$. If $F : C \to \text{Set}$ is a cosheaf, then in particular it sends tensor products to disjoint unions, sends $\mathbb{Z}$ to $\emptyset$, and preserves epimorphisms. Hence it sends $\mathbb{F}_p$ for all primes $p$ to $\emptyset$. Hence it sends $\mathbb{F}_p \otimes \mathbb{F}_q \cong 0$ (where $p \neq q$) to $\emptyset$. And now it follows as above that $F$ is the trivial cosheaf.

No. Let $C$ be any category with a zero object. If $F : C \to \text{Set}$ is a cosheaf, then in particular it sends the zero object to the empty set. But since the zero object is also the terminal object, every object in $c$ is equipped with a morphism $c \to 0$, from which it follows that $F(c)$ is equipped with a morphism to the empty set. Hence $F(c)$ is itself empty, so $F$ is the trivial cosheaf. Now take $C$ to be, say, the category of finite-dimensional vector spaces.

No. Let $C$ be any category with a zero object. If $F : C \to \text{Set}$ is a cosheaf, then in particular it sends the zero object to the empty set. But since the zero object is also the terminal object, every object in $c$ is equipped with a morphism $c \to 0$, from which it follows that $F(c)$ is equipped with a morphism to the empty set. Hence $F(c)$ is itself empty, so $F$ is the trivial cosheaf. Now take $C$ to be, say, the category of finite-dimensional vector spaces.

Edit: In fact the free example suffices. Let $C$ be the free category on a pair of parallel morphisms $f, g : c \to d$. Then the coproduct $d \sqcup d$ exists and both inclusions $d \to d \sqcup d$ are isomorphisms. It follows that if $F$ is a cosheaf then $F(d)$ is the empty set, from which it follows that $F(c)$ must also be the empty set and $F(f) = F(g)$ must be the unique morphism $\emptyset \to \emptyset$, so $F$ is again the trivial cosheaf.

No. Consider the poset of subsets of a $3$-element set $\{ 0, 1, 2 \}$ regarded as a category $C$. I claim there are no cosheaves on $C$ whatsoever. To see this, let $F : C \to \text{Set}$ be a functor. Then $F( \{ 1 \} ), F( \{ 2 \} ), F( \{ 3 \})$ are three sets $P, Q, R$. If $F$ preserves coproducts, then it sends unions to disjoint unions, so we must have $F( \{ 1, 2 \} ) = P \sqcup Q, F( \{ 2, 3 \} ) = Q \sqcup R, F( \{ 3, 1 \}) = R \sqcup P$, and furthermore $F( \{ 1, 2, 3 \} ) = P \sqcup Q \sqcup R$, but the natural map from the disjoint union of $P \sqcup Q$ and $Q \sqcup R$ to this set is not an isomorphism; contradiction. So $F$ cannot preserve coproducts after all.

To get an example with a nontrivial pair of parallel morphisms just add a pair of parallel morphisms to $C$.

No. Let $C$ be any category with a zero object. If $F : C \to \text{Set}$ is a cosheaf, then in particular it sends the zero object to the empty set. But since the zero object is also the terminal object, every object in $c$ is equipped with a morphism $c \to 0$, from which it follows that $F(c)$ is equipped with a morphism to the empty set. Hence $F(c)$ is itself empty, so $F$ is the trivial cosheaf. Now take $C$ to be, say, the category of finite-dimensional vector spaces.