A property of the reject of modules

Question

Let $M$ and $X$ be two finitely generated modules over a finite dimensional algebra $A$. The reject of $M$ in $X$ is $$rej_M(X)=\cap_{f:X \rightarrow M} Ker f.$$

I have seen in a place that if $Hom_A(X,M) \not =0$, then there is a monomorphism $X/rej_{M}(X) \rightarrow M'$ with $M' \in addM$.

I am confused with this result. In my mind, $X/ rej_{M}(X)$ could be cogenerated by $M$, that means there is a monomorphism $X/rej_M(X) \rightarrow M''$ where $M''$ is a direct proudct of $M$. However, $addM$ is the category of the direct sums of the direct summands of $M$. How to get the monomorphism $X/rej_{M}(X) \rightarrow M'$ with $M' \in addM$?

@ Rafael Holanda It is a full subcategory of $mod A$ consisting of direct sums of direct summands of $M$ — Xiaosong Peng, Sep 19 '18 at 03:34

score 1 · Accepted Answer · answered Sep 19 '18 at 03:11

The key point is that $rej_M(X)$ must actually be the intersection of finitely many kernels of homomorphism $X\to M$. This is because $X$ is finite dimensional and hence artinian: any decreasing sequence of submodules of $X$ stabilizes. So, when constructing a monomorphism $X/rej_M(X) \rightarrow M''$, we can take $M''$ to be a product of finitely many copies of $M$, which is also a direct sum of copies of $M$.

A property of the reject of modules

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