I am given the following commutative diagram with non-split exact rows with $L, M, N$ being finite dimensional $A$ modules where $A$ is a finite dimensional $\mathbb{C}$ algebra.
where $L$ is an indecomposable module.
The claim is : If $w$ is an automorphism then so is $v$ and $u$.
This is Lemma 5.4.7. in the book An Introduction to Quiver Representations- Derksen.
The author claims that by taking $w^{-1}$ and considering the appropriate fibered product we can assume that $w = 1_{N}$.
I don't understand this.
Basically he is saying it is enough to prove the claim for $w = I_{N}$.
My attempt: Assume the result is true for $w = I_{N}$. Then we get that for any $v$ such that $g \circ v = g$ implies $v$ is an automorphism (This is also an isolated condition called $g$ is right minimal map ).
But how to use this to show that for any $w$ if there is a $v$ such that $w \circ g = g \circ v$ then $v$ is an automorphism? I dont see how to write this condition in the above form: $g$ composed with something on the right equals $g$.
Please help.
This should just be the argument from the remainder of their proof in the book:
Assuming that $w = 1_N$, we want to show first that $u$ is an automorphism. Towards a contradiction, assume that $u$ is not. Then by Fitting's Lemma, $u$ is nilpotent, so $u^m = 0$ for some $m>0$. Applying $f\circ u = v\circ f$ several times, we also get that $0 = f \circ u^m = v^m \circ f$.
Next, as they argue, by the universal property of the cokernel of the map $f$, the morphism $v^m \colon M \to M$ factors through $\mathrm{coker} f$, which happens to be $N$. Hence, there is $h \colon M\to N$ such that $v^m = h \circ \mathrm{Coker} f = h \circ g$.
Now, use $g \circ v = w \circ g = w$ to show that $ghg = g$. Finally, since $g$ is an epimorphism, right cancellation gives $gh = 1_N$, implying that the exact sequence splits. This is the contradiction we were after.