# Why is a linear representation of a quiver a functor?

Let $$Q:=(Q_0,Q_1,s,t)$$ be a quiver and $$k$$ some field. A representation $$(M,\rho)$$ of $$Q$$ over $$k$$ is the following data:

1. A $$k$$-vector space $$M_v$$ for every $$v\in Q_0$$;
2. A $$k$$-linear map $$\rho(a):M_{s(a)}\rightarrow M_{t(a)}$$, for every $$a\in Q_1$$.

I read that we can see a representation as a functor from the free category $$\text{FrCat}(Q)$$ on $$Q$$. I'm not clear how this is possible because $$\text{FrCat}(Q)$$ usually has more arrows than $$Q$$.

Lets take an example. Consider the following quiver and one of its possible representations: $$Q:=x\xrightarrow{a}y\xrightarrow{b}z\quad \text{and}\quad M:=k^{\oplus 2}\xrightarrow{[1\,0]}k\xrightarrow{\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}}k^{\oplus 3}.$$ The objects of the free category are $$x,y,z$$. The homsets are as follows: \begin{align*} \text{Hom}(x,x)&:=\{1_x\},\\ \text{Hom}(y,y)&:=\{1_y\},\\ \text{Hom}(z,z)&:=\{1_z\},\\ \text{Hom}(x,y)&:=\{(x;a;y)\},\\ \text{Hom}(x,z)&:=\{(x;a,b;z)\},\\ \text{Hom}(y,x)&:=\emptyset,\\ \text{Hom}(y,z)&:=\{(y;b;z)\},\\ \text{Hom}(z,x)&:=\emptyset,\\ \text{Hom}(z,y)&:=\emptyset. \end{align*} Clearly the free category on $$Q$$ has more arrows than the quiver $$Q$$. This makes it unclear how one has to interpret the representation as a functor. The data of our representation is as follows:

• $$M_x:=k^{\oplus 2}$$, $$M_y:=k$$ and $$M_z:=k^{\oplus 3}$$.
• $$\rho(a):=\begin{bmatrix} 1 & 0 \end{bmatrix}$$ and $$\rho(b)=\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}.$$

Here are two problems. Any functor from $$\text{FrCat}(Q)$$ to $$\textbf{Vect}_k$$ would have to specify: (i) an image, e.g., for $$1_x$$, but the representation data has no such image since the original quiver doesn't have a loop on $$x$$; (ii) an image for the path $$(x;a,b;z)$$ which respects composition, but the representation data has no such image since $$x\rightarrow z$$ is not an arrow in $$Q$$ (it's an arrow only in $$\text{FrCat}(Q))$$.

So it seems that any such functor would in fact “misrepresent” the original quiver by adding data which the quiver does not possess. But then how can the representation be correctly described as a functor?

• There is actually no extra data, because there is no choice – it is forced by the rules. For example, identities must go to identities. Dec 30, 2022 at 22:35

Functors must be compatible with identities and with composition of morphisms. So if $$Q$$ is any quiver and $$F$$ is a functor from $$\mathrm{FrCat}(Q)$$ into another category $$\mathcal{C}$$, then

• $$F(1_x) = 1_{F(x)}$$ for every vertex $$x$$ of $$Q$$, and

• for every path $$(x; a_1, a_2, \dotsc, a_n; z)$$ in $$Q$$ we have \begin{align*} F( (x; a_1, a_2, \dotsc, a_n; z) ) &= F\Bigl( (y_{n-1}; a_n; z) ∘ \dotsb ∘ (y_1; a_2; y_2) ∘ (x; a_1; y_1) \Bigr) \\ &= F((y_{n-1}; a_n; z)) ∘ \dotsb ∘ F((y_1; a_2; y_2)) ∘ F((x; a_1; y_1)) \,. \end{align*}

The functor $$F$$ is therefore uniquely determined by its actions on the vertices and arrows of $$Q$$. The category $$\mathrm{FrCat}(Q)$$ has, in fact, the following universal property:

Let $$Q$$ be a quiver and let $$\mathcal{C}$$ be a category. Every homomorphism of quivers from $$Q$$ to (the underlying quiver of) $$\mathcal{C}$$ extends uniquely to a functor from $$\mathrm{FrCat}(Q)$$ to $$\mathcal{C}$$.

It is precisely because of this universal property that $$\mathrm{FrCat}(Q)$$ is called the free category on $$Q$$.

In your specific example, the functor $$F$$ corresponding to the representation $$ρ$$ is given on objects by $$F(x) = M_x = k^{⊕2} \,, \quad F(y) = M_y = k \,, \quad F(z) = M_z = k^{⊕3} \,,$$ and on morphisms by \begin{gather*} F(1_x) = \mathrm{id}_{k^{⊕2}} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \,, \quad F(1_y) = \mathrm{id}_k = \begin{bmatrix} 1 \end{bmatrix} \,, \quad F(1_z) = \mathrm{id}_{k^{⊕3}} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \,, \\[1em] F( (x; a; y) ) = ρ(a) = \begin{bmatrix} 1 & 0 \end{bmatrix} \,, \quad F( (y; b; z) ) = ρ(b) = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \,, \\[1em] \begin{aligned} F( (x; a, b; z) ) &= F( (y; b; z) ∘ (x; a; y) ) \\[0.3em] &= F( (y; b; z) ) ∘ F( (x; a; y) ) \\[0.3em] &= ρ(b) ∘ ρ(a) \\[0.3em] &= \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \begin{bmatrix} 1 & 0 \end{bmatrix} \\[0.3em] &= \begin{bmatrix} 1 & 0 \\ 0 & 0 \\ 0 & 0 \end{bmatrix} \,. \end{aligned} \end{gather*}