I am running the following example in GAP using the QPA package:
Q := Quiver(5, [ [1,2,"a1"], [2,1,"a2"], [2,3,"b"], [3,4,"c"], [4,5,"d"], [5,1,"e"] ]);
kQ := PathAlgebra(Rationals, Q);
AssignGeneratorVariables(kQ);
relations := [a1*b*c, a2*a1, b*c*d*e];
A := kQ/relations;
projectives := IndecProjectiveModules(A);
simples := SimpleModules(A);
Display(Dimension(Hom(Rationals, projectives[1], simples[2])));
and at the end it displays $4$ as the dimension. This is confusing me because of the following theorem (from Quiver Representations by Schiffler, pg. 44):
Theorem 2.11: let $M=(M_i, \psi_\alpha)$ be a representation of $Q$. Then for any vertex $i$ in $Q$, there is an isomorphism of vector spaces $$\operatorname{Hom}(P(i),M)\cong M_i$$
where $P(i)$ is the projective module/representation at vertex $i$, and $M_i$'s are the component vector spaces of the representation $M$.
So in my above example, I use $M=S(2)$ as the simple representation/module at vertex $2$ and $i=1$. So the hom-space should be isomorphic to the first vector space of $S(2)$, which is the zero space, so the dimension I get at the end should be zero.
Question: why am I getting dimension $4$ at the end of the above code rather than $0$. I see no need for the ground field to be algebraically closed in the proof of the above theorem, so I don't think using $\Bbb Q$ is the problem.
The command $\text{Hom}$ is a basic GAP command that takes two vector spaces and constructs the space of linear maps between them. Here it seems to be regarding $P(1)$ and $S(2)$ as vector spaces, four-dimensional and one-dimensional respectively, so the space of linear maps is four-dimensional.
The package QPA has a command $\text{HomOverAlgebra}$ that gives a basis of the space of module homomorphisms (in the case of $P(1)$ and $S(2)$, the zero space).