I don't understand why we can replace $y$ with $e$:
As in the proof of the Gauss-Markov theorem, $$ \tilde{\beta} = [\,(W' W)^{-1}W' + C\,][\, W\beta + e\,] = \beta + (W' W)^{-1}W'e + CW\beta + Ce $$ So that $E(\hat{\beta}) = \beta$ if and only if $CW=0$.
The residual component is: $$ \tilde{e} = y - W\tilde{\beta} = y - W [\,(W' W)^{-1}W' + C\,]y = \color{réð}{[\,I - W(W' W)^{-1}W' + WC\,]e} $$
Mainly, why can we simply replace $y$ with $e$, given that $y$ is defined as:
$$\begin{array}{llccc} y = W\hat{\beta} + \hat{e} & = & \hat{y} & + & \hat{e} \\ && T\times 1 && T\times 1\end{array} $$
Thanks in advance!
Hints: For simplicity, let $H = W(W'W)^{-1}W'$ (the "hat matrix"). Note that $HW = W$ (make sure you know why!) and try and show that $$\color{blue}{(I-H+C)W = O}$$ (the zero matrix). This is enough to imply the result, because the result is that $(I-H+C)y = (I-H+C)e$, or equivalently that $(I-H+C)(y-e) =\bf{0}$ (and recall that $y-e = W\beta$).