$\gamma=(Id\times T)_{\#}\mu\implies(x,y)=(x,T(x))$ $\;\gamma$-a.e. Suppose a transport plan $\gamma\in\Pi(\mu,\nu)$ is induced by a map $T$, i.e. $\gamma=(Id\times T)_{\#}\mu$. I would like to show that $(x,y)=(x,T(x))$ $\;\gamma$-a.e.
My attempt : Consider $(x,y)\notin graph(T)$. If I could take a ball $B((x,y),\epsilon)$ such that $B((x,y),\epsilon)\cap graph(T)=\emptyset$ then $\gamma(B((x,y),\epsilon))=\mu(\{x\;|\;(x,T(x))\in B(0,\epsilon))\})=\mu(\emptyset)=0$ meaning that $(x,y)$ is not in the support of $\gamma$
But I am not sure how to justify the step "I could take a ball $B((x,y),\epsilon)$ such that $B((x,y),\epsilon)\cap graph(T)$".
unrolling the definition :
\begin{align}
\gamma(graph(T)^c)&=(Id\times T)_{\#}\mu(graph(T)^c)\\
&=(Id\times T)_{\#}\mu(\{(x,y)\;|\; y\neq T(x)\})\\
&=\mu((Id\times T)^{-1}(\{(x,y)\;|\; y\neq T(x)\}))\\
&=\mu((\{x\;|\; (x,T(x))\in\{(x,y)\;|\; y\neq T(x)\}\}))\\
&=\mu(\emptyset)\\
&=0
\end{align}

Let $G$ denote the graph of $T$. You want to prove that $\gamma(G)=1$.
Note that $$\begin{align}\gamma(G) &= \mu((Id\times T)^{-1}(G)) \\
&=\mu(\{x: (Id\times T)(x) \in G\}) \\
&= \mu(\{x: (x,T(x)) \in G\}) \\
&= \mu(X)=1 .\end{align}$$
