How does control function approach resolve endogeneity? Suppose I want to estimate
$$Y = \beta_1 + \beta_2 X + \varepsilon$$
Now I know that $X$ and $Y$ are also reversely related
$$X = \gamma_1 + \gamma_2 Y + \xi$$
such that $Cov(\varepsilon,X) \neq 0$.
$Cov(\varepsilon,\xi) = 0$ by assumption.
$Cov(\varepsilon,X) \neq 0$ is shown as follows:
$$ \begin{align}
Cov(\varepsilon, X) &= Cov(\varepsilon, \gamma_1 + \gamma_2 Y + \xi) \\
&= Cov(\varepsilon, \gamma_1 ) + Cov(\varepsilon, \gamma_2 Y) + Cov(\varepsilon, \xi) \\
&= \gamma_2 Cov(\varepsilon, Y) \\
&= \gamma_2 Cov(\varepsilon, \beta_1 + \beta_2 X + \varepsilon) \\
&= \gamma_2 Cov(\varepsilon, \beta_1) + \gamma_2 Cov(\varepsilon, \beta_2 X) + \gamma_2 Cov(\varepsilon, \varepsilon) \\
\end{align} $$
where $Cov(\varepsilon, \varepsilon) = Var(\varepsilon) > 0$. Hence $Cov(\varepsilon,X) \neq 0$.
The Control function approach tries to solve endogeneity by dividing $\varepsilon$ into an endogenous part $\beta_3\nu$ and the exogenous part $e$, where $\nu$ comes from the equation
$$X = \delta_1 + \delta_2 Z + \nu$$
with $Z$ a relevant instrumental variable, and $Cov(Z,\varepsilon) = 0$
Our equation for $Y$ then becomes:
$$Y = \beta_1 + \beta_2 X + \beta_3\nu + e$$
In words I understand that, conditional on $\nu$, $\varepsilon$ is independent of $X$.
I fail to see, however, how this ensures that $Cov(e, X) = 0$.
I would like to derive this in much the same way as shown above for $Cov(\varepsilon, X) \neq 0$.
Any help much appreciated.
 A: Recall that when we consider linear projections of the form
$$W = \beta X + \epsilon,$$
by construction $Cov(X,\epsilon) = 0$. This, combined with the intuition that we split $\epsilon$ into an exogenous and endogenous part (as you mention), yields the result.
In particular, given
$$X = \delta_1+\delta_2 Z + \nu,$$
we know that $Cov(Z,\nu) =0$. Furthermore, if we write out the linear projection of $\epsilon$ on $\nu$, we have
$$\epsilon = \beta_3 \nu + e,$$
so that by construction, $Cov(\nu, e) = 0$.
By the exogeneity assumption, we know that $Cov(Z,\epsilon) = 0$, and so we can derive that
$$Cov(Z,e) = Cov(Z, \epsilon - \beta_3\nu) = Cov(Z,\epsilon) - \beta_3 Cov(z,\nu) = 0.$$
This should make intuitive sense (can you put this in words?). So then, to show  $Cov(e,X) = 0$, we use the fact that $X = \delta_1+\delta_2 Z + \nu$ and that we showed $Cov(Z,e) = Cov(\nu,e) = 0$ to then get that
\begin{align}
Cov(X,e) & = Cov(\delta_1+\delta_2 Z + \nu,e) \\
& = Cov(\delta_1,e) + \delta_2Cov(Z,e) + Cov(\nu,e) \\
& = 0,
\end{align}
as required.
