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Let $\Gamma=(\Gamma_0,\Gamma_1)$ be a quiver and $k\Gamma$ the path algebra of $\Gamma$, let $Rep \Gamma$ the category of finite representations of $\Gamma$, I'm reading a proof of the equivalence between this category and the category of $k\Gamma$-modules of finite dimension $f.d(k\Gamma)$.

To show this equivalence, they first construct the functor F from $Rep\Gamma$ to $f.d(k\Gamma)$., for $(V,f)$ in $Rep\Gamma$ they define $F(V,f)$ to be the direct sum of the vector spaces of $V$, that is $\oplus_{i \in \Gamma_0} V(i)$,for each arrow $\alpha:i\rightarrow j$ there is a $k$-linear aplication $f_\alpha: V(i)\rightarrow V(j)$, so they use the induced map

$$ \overline{f_{\alpha}}=\epsilon_{j}f_{\alpha}\pi_{i}:F(V.F)\rightarrow F(V,F)$$ where $\pi_{i}:F(V,f)\rightarrow V(i)$ is the projection, and $\epsilon_{i}:V(i)\rightarrow F(V,f)$ is the inclusion. and for trivial paths

$$ \overline{f_{e_{i}}}=\epsilon_{j}f_{e_i}\pi_{i}:F(V.F)\rightarrow F(V,F)$$, where $f_{e_i}$ is the identity in $V(i)$

Then they say that therefore $\overline{f}:k\Gamma_0\rightarrow End_k(F(V,f))$ is a k-algebra morphism and $\overline{f}:k\Gamma_1\rightarrow End_k(F(V,f))$ a $k\Gamma_0$-bimodule morphism.

I don't understand this last part of the argument, how to define $\overline{f}$ ? and what is $k\Gamma_0$ and $k\Gamma_1$ , how they arrive to this conclusions ?.

Thank you.

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    I think there may have some place not precise. And I think that $\overline{f}: k \Gamma \rightarrow End_k(F(V,f))$ is a k-algebra morphism. To how to define $\overline{f}$, since we have defined $\overline{f}_{\alpha}$ and $\overline{f}_{e_i}$, then for each path, for example $\alpha_1 \alpha_2$, $\overline{f}_{\alpha_1 \alpha_2}$ is just the composition of $\overline{f}_{\alpha_1}$ and $\overline{f}_{\alpha_2}$. – 2017-03-08

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We can regard $k\Gamma_0$ and $k\Gamma_1$ as two vector spaces which consisting of $k$-valued functions on $\Gamma_0$ and $\Gamma_1$, respectively. The space $k\Gamma_1$ is an $\Gamma_0$-bimodule defined as follows:

$(e\cdot f)(a)=e(ha)f(a)$ and $(f\cdot e)(a)=f(a)e(ta) $ for all $a\in \Gamma_1$, where $e\in k\Gamma_0$ , $f\in k\Gamma_1$, $h:\Gamma_1 \rightarrow \Gamma_0$(source) and $t:\Gamma_1 \rightarrow \Gamma_0$(target).