On the commutativity of tensor product in $\mathscr{D}'$

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Citing Wikipedia:

Let $U\subseteq \mathbb {R} ^{m}$ and $V\subseteq \mathbb {R} ^{n}$ be open sets. Assume all vector spaces to be over the field $\mathbb {F}$, where $\mathbb{F} =\mathbb {R}$ or $\mathbb {C}$. For $f\in {\mathcal {D}}(U\times V)$ define for every $u \in U$ and every $v\in V$ the following functions: $$ {\begin{alignedat}{9}f_{u}:\,&V&&\to \,&&\mathbb {F} &&\quad {\text{ and }}\quad &&f^{v}:\,&&U&&\to \,&&\mathbb {F} \\&y&&\mapsto \,&&f(u,y)&&&&&&x&&\mapsto \,&&f(x,v)\\\end{alignedat}}$$ Given $S\in {\mathcal {D}}^{\prime }(U)$ and $T\in {\mathcal {D}}^{\prime }(V)$, define the following functions: $${\begin{alignedat}{9}\langle S,f^{\bullet }\rangle :\,&V&&\to \,&&\mathbb {F} &&\quad {\text{ and }}\quad &&\langle T,f_{\bullet }\rangle :\,&&U&&\to \,&&\mathbb {F} \\&v&&\mapsto \,&&\langle S,f^{v}\rangle &&&&&&u&&\mapsto \,&&\langle T,f_{u}\rangle \\\end{alignedat}}$$ where $\langle T,f_{\bullet }\rangle \in {\mathcal {D}}(U)$ and $\langle S,f^{\bullet }\rangle \in {\mathcal {D}}(V)$. These definitions associate every $S\in {\mathcal {D}}'(U)$ and $T\in {\mathcal {D}}'(V)$ with the (respective) continuous linear map: $${\begin{alignedat}{9}\,&{\mathcal {D}}(U\times V)&&\to \,&&{\mathcal {D}}(V)&&\quad {\text{ and }}\quad &&\,&&{\mathcal {D}}(U\times V)&&\to \,&&{\mathcal {D}}(U)\\&f&&\mapsto \,&&\langle S,f^{\bullet }\rangle &&&&&&f&&\mapsto \,&&\langle T,f_{\bullet }\rangle \\\end{alignedat}}$$ Moreover if either $S$ (resp. $T$) has compact support then it also induces a continuous linear map of $C^{\infty }(U\times V)\to C^{\infty }(V)$ (resp. $C^{\infty }(U\times V)\to C^{\infty }(U)$).

Fubini's theorem for distributions. Let ${\displaystyle S\in {\mathcal {D}}'(U)}$ and ${\displaystyle T\in {\mathcal {D}}'(V)}$. If ${\displaystyle f\in {\mathcal {D}}(U\times V)}$ then $\langle S,\langle T,f_{\bullet }\rangle \rangle =\langle T,\langle S,f^{\bullet }\rangle \rangle.$ The tensor product of ${\displaystyle S\in {\mathcal {D}}'(U)}$ and $T\in {\mathcal {D}}'(V)$, denoted by $S\otimes T$ or $T\otimes S$ is the distribution in $U\times V$ defined by: $${\displaystyle (S\otimes T)(f):=\langle S,\langle T,f_{\bullet }\rangle \rangle =\langle T,\langle S,f^{\bullet }\rangle \rangle}$$

Question. Does that mean the tensor product is commutative? It's not true in general, though

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No, the tensor product of distributions is not commutative.

If we rewrite the last line with formal arguments we have $$ \langle (S \otimes T)(x,y), f(x,y) \rangle = \langle S(x), \langle T(y), f(x,y) \rangle \rangle = \langle T(y), \langle S(x), f(x,y) \rangle \rangle . $$ Had the tensor product been commutative then the last expression would have read $\langle T(x), \langle S(y), f(x,y) \rangle \rangle$ or $\langle T(y), \langle S(x), f(y,x) \rangle \rangle$.