The Ext functor in the quiver representation

First take a question as an example:

Let $f:L\to M$ be an irreducible morphism in $\mathrm{mod}-A$ and $X$ be a right $A$-module.
Show that $\mathrm{Ext}_A(X,f):\mathrm{Ext}_A(X,L)\to\mathrm{Ext}_A(X,M)$ is a monomorphism,if $\mathrm{Hom}_A(M,X)=0$.

What is the special meaning of $\mathrm{Ext}$ functor in the quiver representation?

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As Matt has said in the comment, representation of quivers, and modules over the corresponding path algebra are really the same thing. Since you said you understand what it means to take Ext in homological algebra, i.e. taking Ext of modules over an algebra. So what you want for the answer is just $Rep(kQ/I) \simeq A-mod$. Here $A\cong kQ/I$, or in general, can be taken as Morita equivalent to $kQ/I$ (which is a basic algebra, meaning simple modules are all 1 dimensional).

We prove simply $Rep(Q)\simeq kQ-mod$ here.

If you have $M\in kQ-mod$, to get a representation of $Q$, take $M_x = e_x M$ where $e_x$ is the trivial path (i.e. primitive idempotent) corresponding to $x\in Q_0$.

If you have $V=\bigoplus_{x\in Q_0} V_x\in Rep(Q)$, then take the vector space $M=V$, and define the action of path $x_1\to \cdots \to x_k$ on $V$ (this lies in the path algebra $kQ$) as composition $V \twoheadrightarrow V_{x_1}\cdots V_{x_k} \hookrightarrow V$. This then gives you a module over the algebra $kQ$.

Hope this helps. Being able to think of a module in terms of representation of quiver is exactly why quiver is so useful to study representation theory. One should never be afraid of representation of quiver. IMO, I find it much harder to study modules over arbitrary algebra.

In answer to your additional question about Example VI.3.11 in [ASS]:

The way too see how the module manifest after tilting is probably not so apparent from the text. The main thing is to note that now the elements in your new algebra $B=End_A(T)$ is given by homomorphisms. The good thing of computation with quiver is you can usually do this rather combinatorially (i.e. looking at the arrows). We look at this example in detail: $3\xleftarrow{\beta}2\xleftarrow{\alpha}1$. Projective indecomposables: $$P(1)=\begin{array}{c}1 \\2\\3\end{array}, P(2)=\begin{array}{c} 2 \\ 3\end{array}, P(3)=\begin{array}{c}3\end{array}$$
On the right hand side of the equal is the way we ‘draw’ the module according to its Loewy structure (radical+socle series), (we should in fact draw them as $S(1)$, $S(2)$, $S(3)$, by for simplicity, we will always omit $S(-)$). You can ‘in some sense’ says that for $P(1)$, the $1$ is representing $e_1$, the $2$ representing $\alpha$ and the $3$ representing $\alpha\beta$. A homomophism between modules must map from the top the domain to somewhere in the target module with the same composition factor, the Loewy structure helps us to visualise what kind of map is possible. Now the tilting module:
$$T(1) = 1, T(2)=\begin{array}{c}1\\ 2\\ 3\end{array}, T(3) = 3$$
Now there is no hom from $T(1)$ to $T(2)$ because, by above statement, 1 must maps to 1, so we have the following diagram:
$$ \begin{array}{ccc} 1 &\to & 1 \\ & & 2 \\& & 3\end{array}$$
which does not make sense because you don’t have anything that maps to 2 and 3 in the target. On the other hand, reversing the arrow is a valid homomorphism. Moreover, this map does not factor through $T(3)$, therefore, by denoting this map as $\lambda^o$, on the quiver of $B^{op}$, we have $1\xleftarrow{\lambda^o}2$ (where 1 represents $T(1)$ and so on).
$T(3)$ is a simple and is socle of $T(2)$, so we also have a map, which does not factor through $T(1)$, denote by $\mu^o$. After reversing the arrow (taking the opposite ring), this gets us the quiver of $B$, $1\xrightarrow{\lambda}2\xrightarrow{\mu} 3$. The following diagram suffices to say $\mu^o\lambda^o(=\lambda\mu)=0$:
$$ \begin{array}{ccccc} 1 &\leftarrow & 1 & & \\ & & 2 & &\\& & 3 & \leftarrow & 3\end{array}$$ Now the projective indecomposables of $B$ are
$$ P_B(3) = 3, P_B(2) = \begin{array}{c}2\\ 3\end{array}, P_B(1) = \begin{array}{c}1\\ 2\end{array}$$
Note the primitive idempotents are given by $f_i:T(i)\to T(i)$ (corresponding to the top of these PIMs). The four equations can now be seen:

  1. $Hom_A(T,3)= K-span\{f_3\}\cong S_B(3)$
  2. $Hom_A(T,\begin{array}{c}1\\2\\3\end{array}) = K-span\{f_2,\mu^o\} \cong P_B(2)$
  3. $Hom_A(T,\begin{array}{c} 1 \\ 2\end{array})$: There is only one map, given by canoncial quotienting of $T(3)$ from $T(2)$, which can be visualise as:
    $$ \begin{array}{ccccc} & & 1 & \xrightarrow{f_2} & 1 \\ & & 2 & \to & 2\\ \hline 3 &\xrightarrow{\mu^o} & 3 & \to & 3 \end{array}$$
    Hence this hom group is given by $K$-span$\{ f_2\}$ quotient by $K$-span$\{\mu^o\}$, this corresponds to $P_B(2)$ quotient out by its socle (=$K$-span$\{\mu\}$. Hence giving us $S_B(2)$
  4. $Hom_A(T,1)=K$-span$\{f_1,\lambda^o\}\cong P_B(1)$