Is $[C,K\text{-}\mathrm{Vect}]$ necessarily equivalent to the module category $K[C]\text{-}\mathrm{Mod}$? I just finished a course on the representation theory of quivers and posets. I recently read the n-lab page on category algebras and how group, path, and incidence algebras are all specific instances of the category algebras. In all of these cases, we have an equivalence between the category of functors from the object to vector spaces and the modules of the category object. (The incidence algebra is a little funny, but it’s my understanding that this is just because we’ve defined things backwards in that case.)
I was wondering how generally this holds. Is the category of modules over a category algebra always equivalent to functors of the category to vector spaces? If not how does this fail?
 A: To make sure we're all on the same page let's clearly state the definitions here. Let $C$ be a category and $K$ a field (or more generally a ring). The category algebra $K[C]$ is the $K$-algebra whose underlying vector space is the free vector space on the set of morphisms in $C$, where the product of two morphisms is their composition in $C$ if they are composable and $0$ otherwise. A functor $F : C \to \text{Vect}_K$ determines a left $K[C]$-module whose underlying vector space is the direct sum $\bigoplus_{c \in C} F(c)$ and whose $K[C]$-algebra structure is determined by the action of $F$ on morphisms.

Claim 1: If $C$ has finitely many objects, this construction is an equivalence of categories between the functor category $[C, \text{Vect}_K]$ and the category of left $K[C]$-modules.

This is a nice exercise. Things get trickier in the case that $C$ has infinitely many objects, because in the finite case the unit is the sum $\sum_{c \in C} \text{id}_c$ and this sum does not exist if $C$ has infinitely many objects. You can show, for example, that if $C$ is a discrete category on infinitely many objects, then the functor category $[C, \text{Vect}_K]$ (which is an infinite product of copies of $\text{Vect}_K$) is not the category of modules over any (unital) algebra. (However, it is the category of comodules over a coalgebra.)

In the infinite case we can replace $K[C]$ with a better-behaved object, namely the free $K$-linear category on $C$ (sometimes also denoted $K[C]$, hence the confusion in the other answer). This is the category with the same objects as $C$ but whose homsets are the free $K$-vector spaces on the homsets of $C$, and it is enriched over $K$-vector spaces. This construction has many advantages over the construction of the category algebra: in addition to generalizing correctly to arbitrary (small) categories $C$, it is also a functor, which the construction of the category algebra is not! (This is a nice exercise too.) It has the property that the category of functors $C \to \text{Vect}_K$ is equivalent to the category of $K$-linear functors $K[C] \to \text{Vect}_K$.
This may not seem like it's bought us much, but the significance of moving to the setting of $K$-linear categories is that we can ask when two $K$-linear categories $A, B$ have the property that their categories of functors $[A, \text{Vect}_K], [B, \text{Vect}_K]$ are equivalent ($K$-linearly). Two such categories are said to be Morita equivalent, and we can show that $A$ and $B$ are Morita equivalent iff they have equivalent Karoubi envelopes (also known as Cauchy completions). So, we can further replace $C$ not only by $K[C]$ but by the Karoubi envelope of $K[C]$. And we can hope that this Karoubi envelope has a simpler description than $C$ itself does.
The most interesting application I know of this idea is to the Dold-Kan correspondence. If we take $A$ to be the (opposite of the) free $K$-linear category on the simplex category, then the Dold-Kan correspondence says that we can take $B$ to be the (opposite of the) "free chain complex." And this can be verified by calculating the Cauchy completion of $A$, although admittedly I've never done this calculation myself.
The category algebra itself can be brought into this circle of ideas: you can show that if $C$ has finitely many objects then the free $K$-linear category $K[C]$ is Morita equivalent to the $K$-algebra $\text{End}_{K[C]} \left( \bigoplus_{c \in C} c \right)$ by calculating that they have the same Cauchy completion. Cauchy completion basically "adds back in any missing direct summands," and so the effect of doing that to $\bigoplus_{c \in C} c$, as an object in $K[C]$, is basically to recover all of the individual objects $c \in C$.
A: Yes, these are the same. Stated more formally, $K[C]$ is the free $K$-linear category generated by $C:$ the property you suggest holds for functors from $C$ to any $K$-linear category $D.$ Indeed, since the hom-spaces in $K[C]$ are freely generated by the hom-sets in $C,$ a $K$-linear functor $K[C]\to D$ is uniquely determined by its restriction to $C;$ since the composition in $K[C]$ is induced from $C,$ a $K$-linear operation $K[C]\to D$ is a  functor if and only if its restriction to $C$ is.
Edit: It was pointed out that by $K[C]$ you likely mean the actual algebra generated by morphisms of $C$ with multiplication induced by composition, and non-composable morphisms multiplying to $0.$ From a functor $F:C\to K-\mathrm{Vect}$, we can define a $K[C]$-module for this sense of $K[C]$ with underlying vector space $\oplus_c F(c)$ and a morphism $f:c\to c'$ acting according to $F(f)$ on the $F(c)$ component and otherwise as $0.$ In the other direction, given a $K[C]$-module $V,$ we define $F:C\to K-\mathrm{Vect}$ by sending $c$ to the cokernel of the action of $1_c$ on $V$ and $f$ to the action of $f,$ which descends to the cokernels since $1_c\cdot f\cdot v=f\cdot 1_c\cdot v=0$ if $v$ is in the kernel of $1_c$ on $V.$ If $C$ is finite, then these operations are clearly mutually inverse, but in general they need not be. Note this constraint is necessary for quivers as well, contrary to what you suggest in your post.
