Dimensions of homomorphisms determines the module I am reading this paper about degenerations for representations.
I am mentioning some context though all of it will not be needed I think for my question.
We basically have a finite dimensional algebra $A$(a path algebra) over an algebraically closed field $k$ and we consider finite dimensional $A/I$ modules. We make an algebraic variety out of the collection of all modules and then ask for orbit closures(orbits corresponding to isomorphism classes).
But my question is:
Show that if $A$ is an algebra such that for any non projective $V$ over $A$ there exists an almost split sequence (Auslander-Reiten Sequence) stopping at $V$ then  $V$ is determined upto isomorphism by the numbers $dim_{k}(Hom(U, V))$. where $U$ is varying over indecomposable modules.
This is mentioned in the paper just before corollary 2.3
I have read about these sequences from Derksen's book An Introduction to Quiver Representations but I can't get started with this problem.
Please help.
 A: We'll use the dual of the condition you mentioned: for any non-injective indecomposable $A$-module $V$, there exists an almost split sequence $0\to V \to E_V \to \tau^{-1} V \to 0$.
Let $M$ and $N$ be finite-dimensional $A$-modules such that for any indecomposable module $U$, the dimensions of $Hom_A(U,M)$ and $Hom_A(U,N)$ are equal.  Let us show that $M$ and $N$ are isomorphic.  To do this, we will use that a decomposition of a finite-dimensional module into a direct sum of indecomposable modules is unique up to isomorphism and reordering of the direct factors (Krull--Schmidt theorem).
Let $U$ be a non-injective indecomposable module.  Applying the functor $Hom_A(-,M)$ to the almost split sequence starting in $U$, we get an exact sequence of vector spaces
$$
0\to Hom_A(\tau^{-1}U, M) \to Hom_A(E_U, M) \to Hom_A(U, M) \to C_{U,M} \to 0,
$$
where $C_{U,M}$ is a vector space whose dimension is the multiplicity of $U$ in a decomposition of $M$ into direct factors -- this follows from the definition of an alsmost split sequence.   Thus the multiplicity of $U$ in a decomposition of $M$ is $$\dim C_{U,M} = \dim Hom_A(\tau^{-1}U, M) - \dim  Hom_A(E_U, M) +\dim  Hom_A(U, M).$$
Replacing $M$ with $N$ in the above, we can also compute the multiplicity of $U$ in a decomposition of $N$: $$\dim C_{U,N} = \dim Hom_A(\tau^{-1}U, N) - \dim  Hom_A(E_U, N) +\dim  Hom_A(U, N).$$  By our assumption on $M$ and $N$, the right-hand side of both equations are equal.  Therefore, $M$ and $N$ have the same non-injective indecomposable factors.
It remains to be seen that the indecomposable injective factors of $M$ and $N$ are the same.  By the above, we can write $M\cong X\oplus I_M$ and $N \cong X\oplus I_N$ with $I_M$ and $I_N$ injective and $X$ having no injective direct summand.  Then the vector spaces $Hom_A(U,I_M)$ and $Hom_A(U,I_N)$ have the same dimension for any indecomposable $U$.  Let $S_1, \ldots, S_n$ be the complete list of simple modules (up to isomoprhisms), and let $I_1, \ldots, I_n$ be their injective envelopes.  Then any indecomposable injective is isomorphic to one of the $I_i$, and we have that $\dim Hom_A(S_i, I_j) = \delta_{ij}$.  Thus $\dim Hom_A(S_i, I_M)$ is the multiplicity of $I_i$ in $I_M$; by our assumption on $M$ and $N$, it is the same as the multiplicity of $I_i$ in $I_N$.  This finishes the proof.
