I'm in trouble with the definition of generator of $R\text{-}\mathbf{mod}$ category of left $R\text{-}\mathbf{mod}$. I've found a lot of equivalent definitions. I'm using the one of Jacobson's Basic Algebra, so $X$ is a generator if every module $M$ is sum of submodules all of which are homomorphic images of $X$.
Does it mean that these submodules are images of $X$ via homomorphisms?
Then I've proved that the following are equivalent:
1) $X$ is a generator;
2) the functor $\operatorname{hom}(X, -)$ is faithful;
3) $T(X)=R$ where $T(X) = \sum_{h \in \operatorname{hom}(X,R)} h(X)$;
4) $R$ is a homomorphic image of $X^n$ for some $n$.
Now I need to use the fact that $X$ is a generator if and only if every left $R$-module $M$ is an epimorphic image of some direct sum $\bigoplus_i X$. Is this fact only a different way to write the definition I've used? I don't know what an epimorphic image is and I'm confused about this sentence.
If anybody can help me this will be very appreciated! Thanks!
If $R$ is a unital ring, you know that every module $M$ can be described as a homomorphic image of some free module $F$; since $$ F \cong \bigoplus_{i \in I}R $$ for some index set $I$, you are pretty much done. Explicitly, let $M$ be a left $R$-module and let $F$ be a free module chosen such that you can realize $M$ as the homomorphic image of some map $\phi:F \to M$, i.e., $\operatorname{Image}(\phi) \cong M$. Then since there exists an $n \in \mathbb{N}$ such that $R$ is the homomorphic image of $X^{n}$, you get that $$ F \cong \bigoplus_{i \in I} R \cong \bigoplus_{i \in I} X^n \cong \bigoplus_{i \in I}\left(\bigoplus_{k=1}^{n}X\right) \cong \bigoplus_{j \in J} X $$ where $I$ is some index set and $J$ is the evident new index set given by reindexing the double direct sum. Call the isomorphism from $F$ to this direct sum above $\alpha$, i.e., $$ \alpha:\bigoplus_{j \in J} X \to F. $$ Then pre-composing $\phi$ with $\alpha$ gives a surjection $$ (\phi\circ \alpha):\bigoplus_{j \in J} X \to M $$ where $$ \operatorname{Image}(\phi \circ \alpha) = \operatorname{Image}(\phi) = M $$ because $\alpha$ is an isomorphism and $\phi$ is a surjection.