2SLS and Control Functions (identity of estimator) I would like to show the identity of the 2 Stage Least Squares estimator and the control function estimator. 
Assume a linear regression model
$$y = X\beta + u$$ where $X =[X_1 \ X_2]$ is $n \times k$ and where $X_2$ is endogenous ($\mathbb E[x_{2i}u_i]\not =0$). Let $Z$ be a $n \times l$ matrix $l\geq k$ of instruments (that includes $X_1$ which is assumed to exogenous $\mathbb E[x_{1i}u_i]=0$) and 
$X = Z\Gamma + V$ 
where $\Gamma:=\mathbb E[z_iz_i^\top]^{-1}\mathbb E[z_i y_i]$ such that by construction $\mathbb E[z_iv_i^\top]=0$
The 2SLS Estimator
The two stage least squares estimator is defined as $\hat \beta_{2SLS}:=(\hat X^\top X)^{-1}(\hat X^\top y)$ where $\hat X = Z(Z^\top Z)^{-1} Z X = P_Z X$.
The control function approach regress $X$ on $Z$ to get residuals $\hat V$
$X = Z\hat \Gamma + \hat V$  and then includes these residuals in a regresssion of $y$ on $X$ and $\hat V$ to get the estimator
$$ \begin{bmatrix}\hat b  \\ \hat \rho \end{bmatrix} = \begin{bmatrix} X^\top X & X^\top \hat V \\ \hat V^\top X & \hat V^\top \hat V\end{bmatrix}^{-1} \begin{bmatrix} X^\top y \\ \hat V^\top y\end{bmatrix} $$
where the result I am looking for then is that $\hat b = \hat \beta_{2SLS}$.
 A: When we regress the regressors $Z$ on the instruments $X$, we 1. get residuals $\widetilde{Z}:=M_{X}Z=Z-X(X'X)^{-1}X'Z$.
We then, 2., regress $y$ on $Z$ and $\widetilde{Z}$,
\begin{equation}
y=Z\widehat{\delta}+\widetilde{Z}\widehat{\theta}+\widehat{u}
\end{equation}
Recall that the Frisch-Waugh-Lovell theorem states that we can obtain subvectors of coefficients on variables of "interest" of a long regression by regressing the residuals of a regression of the dependent variable on the remaining ("non-interesting'') explanatory variables on the residuals of a regression of the submatrix of interest on the remaining variables.
We use FWL in step 2 to show that
\begin{eqnarray*}
\widehat{\delta}&=&(Z'M_{\widetilde{Z}}Z)^{-1}Z'M_{\widetilde{Z}}y\\
&=&(Z'(I-P_{\widetilde{Z}})Z)^{-1}Z'(I-P_{\widetilde{Z}})y
\end{eqnarray*}
Now,
\begin{eqnarray*}
P_{\widetilde{Z}}&=&M_{X}Z(Z'M_{X}'M_{X}Z)^{-1}Z'M_{X}\\
&=&M_{X}Z(Z'M_{X}Z)^{-1}Z'M_{X}
\end{eqnarray*}
so that
$$
(I-P_{\widetilde{Z}})Z=Z-M_{X}Z(Z'M_{X}Z)^{-1}Z'M_{X}Z=Z-M_{X}Z=P_{X}Z
$$
such that
$$
\widehat{\delta}=(Z'P_{X}Z)^{-1}Z'P_{X}y
$$
