There is the well-known construction of sheaves on a base, i.e. rather than specifying a sheaf $S$ on all open sets of a topological space $M$, specify its data only on a topological base $\mathcal{B}$ of $M$, where $\mathcal{B}$ is a collection of open sets in $M$ so that every open set is a union of elements in $\mathcal{B}$. See for example these notes for details.
Now, pick your favourite definition of a cosheaf $P$ on $M$, e.g. a colimit-preserving covariant functor $\text{Open}(M) \to \text{Ab}$ from the category of open sets of $M$ to the category of abelian groups. Is there a good dual notion of a cosheaf on a base? I.e. if one specifies cosheaf data on a base of $M$, does this specify a unique cosheaf on $M$?
Intuitively, it feels very plausible that the dual proof holds up, and I think one at least easily gets the existence of a precosheaf extending the given data.
However, I am aware that a lot of constructions that work for sheaves break down for cosheaves, e.g. cosheafification seems to be a very troubled notion, because colimits somehow behave differently than limits. I do not have good intuition for category theoretic limits and colimits and how they differ, and literature on cosheaf theory seems scarce, which is why I am asking this question. I hope I'm not being too imprecise here 🙂
Best Answer
By definition, a cosheaf on a space $X$ with values in a category $\newcommand\C{\mathcal{C}}\C$ is a sheaf with values in $\newcommand\op{\text{op}}\C^\op$. Thus to understand cosheaves, it suffices to understand sheaves.
In particular, to address your specific question, we have the following result.
Let $B$ be a base for the topology on $X$. Define the category of sheaves on $B$ in the usual way. Recall that sheaves on $B$ are functors $F$ from the opposite of the poset category of $B$ to $\C$ such that $$F(U)= \lim_{V\subseteq U} F(V),$$ where $U\in B$ and $V$ runs across basic open subsets of $U$. Note that I've written $=$ because $F(U)$ comes with a canonical cone given by the restriction maps.
Proof.
We need to show that the restriction functor is always fully-faithful and is essentially surjective if $\C$ is complete.
Let $F,G$ be sheaves on $X$. Let $F_B$, $G_B$ denote the restriction of $F$ and $G$ to the basis $B$. We know that if $U$ is any open subset in $X$, then since $F$ is a sheaf, $$F(U) = \lim_{V\subseteq U, V\in B} F_B(V).$$ So if $\phi :F_B\to G_B$, then $\phi$ induces maps $F(U)\to G(U)$ for all open sets $U$ subset of $X$ compatible with the restriction maps. In other words, $\phi$ extends to a morphism $\phi' : F\to G$ which restricts to $\phi$ (not hard to check) on $B$. Moreover, this extension is unique, by the universal property of the limit.
This proves that the restriction functor is fully-faithful. (Full because all morphisms of $B$-sheaves can be extended, and faithful because the extension is unique).
Now if $\C$ is complete, if $F_B$ is a sheaf on $B$, then we can define $$F(U) = \lim_{V\subseteq U, V\in B} F_B(V),$$ (we need completeness to guarantee that the limit exists), and you can check that this defines a sheaf and $F(V)=F_B(V)$ when $V\in V$. Thus the restriction of $F$ to $B$ is (canonically isomorphic to) $F_B$. Therefore the restriction functor is essentially surjective if $\C$ is complete, and thus an equivalence of categories. $\blacksquare$
In particular, when we want cosheaves valued in abelian groups, these are the same as sheaves valued in $\mathbf{Ab}^\op$, and $\newcommand\Ab{\mathbf{Ab}}\Ab$ is cocomplete, so $\Ab^\op$ is complete. Thus this proposition applies. Cosheaves on a space are equivalent to cosheaves on a basis for that space.