Why is a morphism of quiver representations a subset of this product of morphisms?

Question

Given two representations $M,N$ of a quiver Q, why is the set of all morphisms of representations from $M$ to $N$ a subspace of $\prod_{i\in Q_0} \mathrm{Hom}(M,N)$? I mean why the product of $\mathrm{Hom}$?

Answer 1

Let $Q$ be a quiver consisting of a set of vertices $Q_0$ and a set of edges $Q_1$. Let $R$ be some ring.

A representation $M$ of the quiver $Q$ assigns to each vertex $i$ an $R$-module $M_i$, and to each edge from $i$ to $j$ an $R$-module homomorphism $\varphi_{ij}:M_i\to M_j$.

Now, let $N$ be another quiver representation of $Q$. A morphism $F:M\to N$ of quiver representations is a collection of $R$-module homomorphisms $F_i:M_i\to N_i$, such that the appropriate diagrams commute. That is, if there is an edge between vertices $i$ and $j$, then the representation morphism should satisfy $\psi_{ij}\circ F_i= F_j\circ \varphi_{ij}$, where $\psi_{ij}$ is the morphism between $N_i$ and $N_j$.

We can conclude that a morphism of representations of the quiver $Q$ consists of a morphism $M_i\to N_i$ for each $i$, satisfying some extra conditions. Hence, it can be interpreted as an element in the product $\prod_{i\in Q_0}Hom_R(M_i,N_i)$, satisfying some extra properties.

In particular, if $Q$ is a quiver with no edges, then the set of homomorphisms between two representations $M$ and $N$ is precisely the product $\prod_{i\in Q_0}Hom_R(M_i,N_i)$.

Answer 2

To say what EBP's answer says but with different words, if your quiver $Q$ has vertices $(1,2,\dotsc,n)$, then your representation $M$ has vector spaces $(M_1, M_2, \dotsc, M_n)$ assigned to each node. Then a morphism of representations $M \to N$ is just a bunch of morphisms on these individual vector spaces $(M_1 \to N_1, M_2 \to N_2, \dotsc, M_n \to N_n)$ that satisfy the commutativity relation mentioned. Notice that this is an $n$-tuple of morphisms. The correct way to express the collection of these $n$-tuples is as a product.

$$(M_1 \to N_1, M_2 \to N_2, \dotsc, M_n \to N_n) \in \prod_{i \in Q_0}\mathrm{Hom}(M_i, N_i)$$