Sign issue when computing identity involving curvature for a connection $\nabla + A$ of a bundle $E\to M$ I'm trying to prove the well-known formula which relates the curvature associated to the connection $\nabla$ to the one associated to $\nabla + A$, namely: $$F_{\nabla +A} = F_{\nabla}+d_{\nabla}A+A\wedge A$$ where $\nabla$ is a fixed connection on the vector bundle $E\to M$, and $A\in \Omega^1(End(E))$. These are my computations:
\begin{align}
F_{\nabla + A} \otimes s &= d_{\nabla+A}d_{\nabla+A}s\\
&=d_{\nabla+A}\left(\nabla+A \right)s\\
&=d_{\nabla+A}\left(\nabla s\right) + d_{\nabla+A}\left(A s\right)\\
&= -\omega \wedge \nabla s' -\omega\wedge As'+d\omega\otimes s - \eta \wedge \nabla a(s) - \eta \wedge A(a(s))+d\eta\otimes a(s) \\
&= F_{\nabla}\otimes s  -\omega\wedge As' - \eta \wedge \nabla a(s) - \eta \wedge A(a(s))+d\eta\otimes a(s)\\
&= F_{\nabla}\otimes s  -\omega\wedge As' - \eta \wedge \nabla a(s) - \eta \wedge A(a(s))+d\eta\otimes a(s)\\
\end{align}
Where in the 4th passage I used the fact that $\nabla s =\omega \otimes s'$ and $A(s)=\eta\otimes a(s)$, for $1$-forms $\omega, \eta \in \Omega^1$ and $a\in \text{End}(E)$.
Notice now that 
\begin{align*}
\left( d_{\nabla} A\right)s&= d_{\nabla}(As)-Ad_{\nabla}(s)\\
&=-\eta \wedge \nabla a(s)+d\eta\times a(s)-A\nabla s
\end{align*}
Therefore we have 
\begin{align}
F_{\nabla + A} \otimes s &= F_{\nabla}\otimes s  -\omega\wedge As' - \eta \wedge \nabla a(s) - \eta \wedge A(a(s))+d\eta\otimes a(s)\\
&=F_{\nabla}\otimes s + \left( d_{\nabla} A\right)s - \eta \wedge A(a(s))
\end{align}

I'm tempted to identify $- \eta \wedge A(a(s))$ with $-A\wedge A$, but this means I've a sign issue.

Can someone clarify this to me? Am I making some silly mistakes? 
 A: I can't follow your computations, mostly because I don't understand the types of the objects you are working with. For example, $d_{\nabla + A}(d_{\nabla + A}(s))$ should be an element of $\Omega^2(M;E)$ while $F_{\nabla + A}$ is an element of $\Omega^2(M;\operatorname{End}(E))$. What does $F_{\nabla + A} \otimes s$ mean? The way to generate an element of $\Omega^2(M;E)$ from an element of $\Omega^2(M;\operatorname{End}(E))$ is by contraction, not by tensor product and the correct identity is
$$ (F_{\nabla + A}(X,Y))(s) = ((d_{\nabla + A}^2)(s))(X,Y) $$
and not what you wrote.
Given $\alpha \in \Omega^1(M;E)$, we have the following formula for $d_{\nabla}(\alpha)$ (similar to the formula for $d\alpha$ of a one-form):
$$ (d_{\nabla}(\alpha))(X,Y) = \nabla_X(\alpha(Y)) - \nabla_Y(\alpha(X)) - \alpha([X,Y]).$$
Using this formula, we can calculate:
$$ (F_{\nabla + A}(X,Y))(s) = ((d_{\nabla + A}^2(s))(X,Y) = (d_{\nabla + A}(d_{\nabla + A}(s)))(X,Y) = (d_{\nabla + A}(\nabla_{*}(s) + (A(*))(s)))(X,Y) \\
= \nabla_X(\nabla_Y(s)) + A(X)(\nabla_Y s) - \nabla_Y(\nabla_X s) - A(Y)(\nabla_X s) - \nabla_{[X,Y]}s \\
\nabla_X(A(Y)s) + A(X)(A(Y)s) - \nabla_Y(A(X)s) - A(Y)(A(X)s) - A([X,Y])(s) \\
= \left( \nabla_X(\nabla_Y s) - \nabla_Y(\nabla_X s) - \nabla_{[X,Y]}(s) \right) + \\
\left( \nabla_X(A(Y)s) - A(Y)(\nabla_X s) - \nabla_Y(A(X)s) + A(X)(\nabla_Y s) - A([X,Y])(s) \right)+ \\
(A(X)(A(Y)s) - A(Y)(A(X)s)). $$
On the other hand,
$$ F_{\nabla}(X,Y)(s) = \nabla_X(\nabla_Y s) - \nabla_Y(\nabla_X s) - \nabla_{[X,Y]}(s), \\
(d_{\nabla}(A)(X,Y))(s) = (\nabla_X(A(Y)))(s) - (\nabla_Y(A(X)))(s) - A([X,Y])s \\ 
= \nabla_X(A(Y)s) - A(Y)(\nabla_X s) - \nabla_Y(A(X)s) + A(X)(\nabla_Y s) - A([X,Y])s, \\
((A \wedge A)(X,Y))(s) = (A(X) \circ A(Y) - A(Y) \circ A(X))(s) = A(X)(A(Y)s) - A(Y)(A(X)s).$$
