Exercise 4.7 in SLP (dynamic programming) Exercise 4.7 (b) : Show that under Assumptions 4.10 and 4.11, $T:H(X) \to H(X)$.

$H(X)$ is the set of continuous and homogeneous of degree one functions and $Tf(x) = \sup_{y \in \Gamma(x)} \{F(x,y) + \beta f(y)\}$. The continuity follows from the theorem of maximum. I am a little confused with showing $Tf$ is homogeneous of degree one. According to the solution,
\begin{equation}
\begin{split}
Tf(\lambda x)&amp; = \sup_{\lambda y \in \Gamma(\lambda x)} \{F(\lambda x,\lambda y) + \beta f(\lambda y)\}\\
&amp; = \lambda \sup_{y \in \Gamma(x)} \{F(x,y) + \beta f(y)\} \\
&amp; = \lambda Tf(x).
\end{split}
\end{equation}
$\lambda$ can be taken out of functions because they are homogeneous of degree one. But, how can we justify the change from $\Gamma(\lambda x)$ to $\Gamma(x)$? More precisely,
\begin{equation}
\begin{split}
Tf(\lambda x)&amp; = \sup_{\lambda y \in \Gamma(\lambda x)} \{F(\lambda x,\lambda y) + \beta f(\lambda y)\}\\
&amp; = \lambda \sup_{\lambda y \in \Gamma(\lambda x)} \{F(x,y) + \beta f(y)\} \\
&amp; = \lambda \sup_{y \in \Gamma( x)} \{F(x,y) + \beta f(y)\} .
\end{split}
\end{equation}
I do understand the second equality, but I do not understand the third equality.
Assumption 4.10 and 4.11 are as follows:
It follows from Assumption 4.10.
Let $\lambda y \in \Gamma(\lambda x)$ be the solution.
Let $\delta = \frac{1}{\lambda}$.
By Assumption 4.10,
$$
\lambda y \in\Gamma(\lambda x) 
$$
implies
$$
\delta \lambda y \in \Gamma(\delta\lambda x)
$$
in other words
$$
 y \in \Gamma(x)
$$