I am not able to show the last step in the proof of the following
Theorem: Suppose $X$ and $Y$ are Polish spaces, $\pi$ is a probability measure on $X \times Y$ and $\mu$ is a probability measure on X. Assume $\pi(A \times Y) = \mu(A)$ for all $A \subset X$ borel set. Let $\Gamma \subset X \times Y$ be a $\pi$ - measurable graph on which $\pi$ is concentrated. Then there exists $T:X \to Y$ s.t. $\pi= (\text{id} \times T)_{\#}\mu$.
I recall that if $\mu$ is a measure on a Polish space $X$ and $f: X \to Y$ is a measurable function, the measure $f_{\#}\mu$ on $Y$ is defined as $$ f_{\#}\mu(B) = \mu (f^{-1}(B)) \text{ for all } B \subset Y \text{ Borel set} $$
Proof
Being $\mu$ a tight regular measure I can find a growing sequence $\{K_k \}_{k \in \mathbb{N}} \subset \Gamma$ of compact sets s.t. $\pi(\Gamma \setminus \cup_k K_k)=0$. Call $C_k = \text{proj}_X (K_k)$ and define $T_k(x)$ s.t. $(x,T_k(x) \in \Gamma$. The set $X \setminus \cup_k C_k$ is $\mu$ negligible and we can define $$ T(x)= \begin{cases} T_k(x) \quad & x \in C_k \\ y_0 \quad &x \in X \setminus \cup_k C_k \end{cases} $$ for a fixed $y_0 \in Y$. This definition is well posed since $C_l \subset C_k$ if $l \le k$. Then it is claimed that $$ (\text{id} \times T)_{\#} \mu = \pi$$
How can I see that?