Given a representation \(V\) of \(\Pi_Q'\), we can form a representation of \(e_0 \Pi_Q' e_0\) by simply taking
\[ V \mapsto e_0 V.\]
To recover the reverse process, from a representation \(V_0\) of \(A_0 = e_0 \Pi_Q' e_0\) to a representation of \(\Pi_Q'\), we can use the tensor product (the obvious base change operation)
\[ \mathrm{Rep}(A_0) \to \mathrm{Rep}(\Pi_Q') \]
\[ V_0 \mapsto \Pi_Q' \otimes_{A_0} V_0 \]
Consider \(Q = A_2\), Let \(V_0\) be a representation of \(A_0 = k[x_{01}x_{10}, x_{01}y_{10}, y_{01}x_{10}, y_{01}y_{10}]\)
If \(V_0 = \langle v_i \rangle\) is a basis of \(V_0\), then we have that
\[ V = \Pi_Q' \otimes_{A_0} V_0 = e_1 V \oplus e_0 V = V_1 \oplus V_0 \]
where \(V_1\) is generated by \(x_{10}V_0 + y_{10}V_0\).
i.e. the vectors \(x_{10}v_i, y_{10}v_i\) certainly generates