I am reading Santambrogio's book on optimal transport, remark 1.19. Let's consider an optimal transport problem between $(X,\mu)$ and $(Y,\nu)$.

(Remark 1.19) ... every time that we know that any optimal $\gamma$ must be induced by a map $T$, then we have uniqueness (of a Kantorovich solution). Indeed, suppose that two different plans $\gamma_1=\gamma_{T_1}, \gamma_2=\gamma_{T_2}$ are optimal: consider $\frac{1}{2}\gamma_1+\frac{1}{2}\gamma_2$, which is optimal as well by convexity. This last transport plan cannot be induced by a map unless $T_1=T_2$ $\mu-a.e.$, which gives a contradiction.

Here an optimal plan $\gamma$ refers to a solution of the Kantorovich problem and an optimal map $T$ refers to a solution of the Monge problem. The remark does not assume any regularity of the spaces.

I could prove the statement assuming that $\gamma(\{(x,y)\in X\times Y:y=T(x)\})=1$, but I could only show that $\gamma$ is concentrated on the closure of the set. From that I could deduce $\gamma(\{(x,y)\in X\times Y:y=T(x)\})=1$ when $T$ is continuous $\mu$-a.e. I am also not sure if optimality matters for the statement. So my questions are:

1. Can we prove or disprove that $\gamma(\{(x,y)\in X\times Y:y=T(x)\})=1$?
2. If the answer for 1. is no, how can we prove Remark 1.19?

===========================================================

Edit: For people who are interested in the same question, I complete the proof for Remark 1.19 here. (Thanks to @gerw for the nice answer.)

Lemma If $\gamma=\gamma_T:=(id, T)_{\#}\mu$, then $\gamma(\{(x,y)\}\in X\times Y:y=T(x))=1$.

Proposition Let $\gamma_1=\gamma_{T_1}$ and $\gamma_2=\gamma_{T_2}$. Then $\gamma=\frac{1}{2}\gamma_1+\frac{1}{2}\gamma_2$ cannot be of the form $\gamma=(id,T)_{\#}\mu$ for any optimal transport map $T$ unless $T_1=T_2$ $\mu-a.e.$.

Proof Suppose $\gamma=(id,T)_{\#}\mu$ for some $T$. Then,

$$ \begin{alignat*}{3} 1=\gamma(y=T(x)) &=&&\gamma(\{y=T(x)\}\cap\{T(x)=T_1(x)\ne T_2(x)\}) \\ & &&+\gamma(\{y=T(x)\}\cap\{T(x)=T_2(x)\ne T_1(x)\})\\ & &&+\gamma(\{y=T(x)\}\cap\{T(x)=T_1(x)= T_2(x)\})\\ & &&+\gamma(\{y=T(x)\}\cap\{T(x)\ne T_1(x), T(x)\ne T_2(x)\})\\ &=&& \frac{1}{2}\gamma_1(\{y=T(x)\}\cap\{T(x)=T_1(x)\ne T_2(x)\})\\ & &&+ \frac{1}{2}\gamma_2(\{y=T(x)\}\cap\{T(x)=T_2(x)\ne T_1(x)\})\\ & &&+ \frac{1}{2}\gamma_1(\{y=T(x)\}\cap\{T(x)=T_1(x)= T_2(x)\})\\ & &&+ \frac{1}{2}\gamma_2(\{y=T(x)\}\cap\{T(x)=T_2(x)= T_1(x)\})\\ &=&& \frac{1}{2}\gamma_1(\{y=T_1(x)\}\cap\{T(x)=T_1(x)\})\\ & &&+\frac{1}{2}\gamma_2(\{y=T_2(x)\}\cap\{T(x)=T_2(x)\})\\ &=&& \frac{1}{2}\gamma_1(T(x)=T_1(x))+\frac{1}{2}\gamma_2(T(x)=T_2(x))\\ &=&& \frac{1}{2}\mu(T(x)=T_1(x))+\frac{1}{2}\mu(T(x)=T_2(x)). \end{alignat*} $$

Thus $\mu(T(x)=T_1(x))=\mu(T(x)=T_2(x))=1$.