Let $R$ be a ring. Suppose that $X$ and $Y$ are left injective $R$-modules and $\theta: X \to Y$ and $\phi:Y \to X$ are $R$-monomorphisms. I want to prove that $X \cong Y$.
This is what I have done so far: Since $X \cong \theta(X) \subseteq Y$ and $Y$ is injective, then $\theta(X)$ is a summand of $Y$; that is, $Y = \theta(X)\oplus T$ for some submodule $T\subseteq Y$. My question is: how can I prove that $T = 0$ ?
I appreciate any help. Thanks in advance.
It's like the Schroeder-Bernstein Theorem, but module-theoretically. Read how that theorem is proved (for example here) and then adapt the proof. Instead of using set differences use the direct summands between the stages.