In "Lectures on gauge theory and integrable systems" of M.Audin, she identifies the space of conections $\mathcal{A}$ on the trivial bundle $G\times S$ ($G$ Lie group, $S$ surface with boundary) with the vector space $\Omega^1(S,\mathfrak{g})$. Then, for each $A\in\mathcal{A}$ define

$$d_A:\Omega^0(S,\mathfrak{g})\rightarrow\Omega^1(S,\mathfrak{g});\alpha\mapsto d\alpha+[A,\alpha]$$

where $[A,\alpha](X)=[A(X),\alpha]$ for tangent vector to $S$.

Similarly $$d_A:\Omega^1(S,\mathfrak{g})\rightarrow\Omega^2(S,\mathfrak{g});\phi\mapsto d\phi+[A,\phi]$$ with $[A,\phi](X,Y)=[A(X),\phi(Y)]-[A(Y),\phi(X)]$.

1. For $\alpha\in\Omega^0(S,\mathfrak{g})$, she claims $d_A\circ d_A(\alpha)=[d_AA,\alpha]$ then she defines the curvature form as $F(A)=d_AA$. Is not wrong this calculation? (I think should be $d_A\circ d_A(\alpha) = [dA+\frac{1}{2}[A,A].\alpha]$).

2. With her definition of $F(A)$ she asks to show that $(T_AF)(\phi)=d_A\phi$, but I couldn't. So, I should be consider $F(A)=dA+\frac{1}{2}[A,A]?$ I like to understand this.

In "Lectures on gauge theory and integrable systems" of M.Audin, she identifies the space of connections $\mathcal{A}$ on the trivial bundle $G\times S$ ($G$ Lie group, $S$ surface with boundary) with the vector space $\Omega^1(S,\mathfrak{g})$. Then, for each $A\in\mathcal{A}$ define

$$d_A:\Omega^0(S,\mathfrak{g})\rightarrow\Omega^1(S,\mathfrak{g});\alpha\mapsto d\alpha+[A,\alpha]$$

where $[A,\alpha](X)=[A(X),\alpha]$ for a vector $X$ tangent to $S$.

Similarly $$d_A:\Omega^1(S,\mathfrak{g})\rightarrow\Omega^2(S,\mathfrak{g});\phi\mapsto d\phi+[A,\phi]$$ with $[A,\phi](X,Y)=[A(X),\phi(Y)]-[A(Y),\phi(X)]$.

1. For $\alpha\in\Omega^0(S,\mathfrak{g})$, she claims $d_A\circ d_A(\alpha)=[d_AA,\alpha]$, then she defines the curvature form as $F(A)=d_AA$. Why does this hold? (I think should be $d_A\circ d_A(\alpha) = [dA+\frac{1}{2}[A,A].\alpha]$).

2. With her definition of $F(A)$ she asks to show that $(T_AF)(\phi)=d_A\phi$, but I couldn't. So, should I consider $F(A)=dA+\frac{1}{2}[A,A]?$ I would like to understand this.

In "Lectures on gauge theory and integrable systems" of M.Audin, she identifies the space of conections $\mathcal{A}$ on the trivial bundle $G\times S$ ($G$ Lie group, $S$ surface with boundary) with the vector space $\Omega^1(S,\mathfrak{g})$. Then, for each $A\in\mathcal{A}$ define  $$d_A:\Omega^0(S,\mathfrak{g})\rightarrow\Omega^1(S,\mathfrak{g});\alpha\mapsto d\alpha+[A,\alpha]$$ where $\left\[A,\alpha\right\](X)=[A(X),\alpha]$ for tangent vector to $S$.

Similarly $$d_A:\Omega^1(S,\mathfrak{g})\rightarrow\Omega^2(S,\mathfrak{g});\phi\mapsto d\phi+[A,\phi]$$ with $\left\[A,\phi\right\](X,Y)=\left\[A(X),\phi(Y)\right\]-\left\[A(Y),\phi(X)\right\]$

1.- For $\alpha\in\Omega^0(S,\mathfrak{g})$, she claims $d_A\circ d_A(\alpha)=[d_AA,\alpha]$ then she defines the curvature form as $F(A)=d_AA$. Is not wrong this calculation? (I think should be $d_A\circ d_A(\alpha)=[dA+\frac{1}{2}[A,A].\alpha]$).

2.-Whit her definition of $F(A)$ she asks to show that $(T_AF)(\phi)=d_A\phi$, but I couldn't. So, I should be consider $F(A)=dA+\frac{1}{2}[A,A]?$ I like to understand this.

In "Lectures on gauge theory and integrable systems" of M.Audin, she identifies the space of conections $\mathcal{A}$ on the trivial bundle $G\times S$ ($G$ Lie group, $S$ surface with boundary) with the vector space $\Omega^1(S,\mathfrak{g})$. Then, for each $A\in\mathcal{A}$ define 

$$d_A:\Omega^0(S,\mathfrak{g})\rightarrow\Omega^1(S,\mathfrak{g});\alpha\mapsto d\alpha+[A,\alpha]$$

where $[A,\alpha](X)=[A(X),\alpha]$ for tangent vector to $S$.

Similarly $$d_A:\Omega^1(S,\mathfrak{g})\rightarrow\Omega^2(S,\mathfrak{g});\phi\mapsto d\phi+[A,\phi]$$ with $[A,\phi](X,Y)=[A(X),\phi(Y)]-[A(Y),\phi(X)]$.

1. For $\alpha\in\Omega^0(S,\mathfrak{g})$, she claims $d_A\circ d_A(\alpha)=[d_AA,\alpha]$ then she defines the curvature form as $F(A)=d_AA$. Is not wrong this calculation? (I think should be $d_A\circ d_A(\alpha) = [dA+\frac{1}{2}[A,A].\alpha]$).

2. With her definition of $F(A)$ she asks to show that $(T_AF)(\phi)=d_A\phi$, but I couldn't. So, I should be consider $F(A)=dA+\frac{1}{2}[A,A]?$ I like to understand this.