Let $R$ be a Henselian Cohen-Macaulay (CM) local ring with maximal ideal $\mathfrak m_R$. For an indecomposable finitely generated maximal Cohen-Macaulay (MCM) $R$-module $M$, the set $\mathfrak S(M)$ contains all non-split short exact sequences of the form $$s:0\to N_s\to E_s\to M\to 0,$$ where $N_s,E_s$ are also finitely generated MCM $R$-modules and $N_s$ is indecomposable.
For $s,t\in \mathfrak S(M)$, write $s\ge t$ if $\exists f\in \mathsf{Hom}_R(N_s,N_t)$ with $\mathsf{Ext}_R^1(M,f)(s)=t$; write $s\simeq t$ if moreover $f$ is an isomorphism. I intend to show that $\ge$ is a partial order on $\mathfrak S(M)$ modulo $\simeq$.
It is clear that $\ge$ is reflexive and transitive, so it suffices to show that $\ge$ is anti-symmetric. Let $s,t\in \mathfrak S(M)$ with $s\ge t$ and $t\ge s$. This means $\exists f:N_s\to N_t$, $\exists g:N_t\to N_s$ with $\mathsf{Ext}_R^1(M,f)(s)=t$ and $\mathsf{Ext}_R^1(M,g)(t)=s$. Letting $h:=g\circ f\in \mathsf{End}_R(N_s)$, we have $\mathsf{Ext}_R^1(M,h)(s)=s$. It is clear that if $h$ is an isomorphism, then $g$ is surjective and $f$ is injective, and by symmetry this will finish the proof. However, I have difficulty showing $h$ is an isomorphism.
Thank you very much for any help!
Fixing $s$, let me drop the suffixes. We have the natural map $\phi:\operatorname{End}(N)\to \operatorname{Ext}^1(M,N)$. $S=R[h]\subset\operatorname{End}(N)$, a commutative subring. Since $R$ is hensel and $S$ is a finite module over $R$, it is a product of local rings. If it is not local, then $S$ has non-trivial idempotents, which is impossible, since $N$ is indecomposable. Thus $S$ is local. The induced map $\phi$ from $S\to \operatorname{Ext}^1(M,N)$ has kernel an ideal (I may have missed something here) of $S$. Now, $h-1$ is in the kernel and the kernel is a proper ideal (since the sequence does not split) says $h-1$ is in the maximal ideal of $S$ and so $h$ is a unit. This is what we wanted to prove.