Why products do not exist in the category of measurable spaces and probability kernels? Consider a category whose objects are measurable spaces and morphisms are probability kernels (also called stochastic kernels or Markov kernels). That is, objects are pairs $(X,\mathbf{A})$ where $\mathbf{A}$ is a $\sigma$-field of subsets of $X$, and morphisms $F\colon(X,\mathbf{A})\to(Y,\mathbf{B})$ are maps $F\colon X\times\mathbf{B}\to [0,1]$ such that


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*$F_x\colon B\mapsto F(x,B)$ is a probability measure for every $x\in X$,

*$F_B\colon x\mapsto F(x,B)$ is a measurable function for every $B\in\mathbf{B}$.


The composition $G\circ F$ of morphisms $F\colon(X,\mathbf{A})\to(Y,\mathbf{B})$ and $G\colon(Y,\mathbf{B})\to(Z,\mathbf{C})$ is defined by
$$(G\circ F)(x,C)=\int_{y\in Y}G_C(y)\,\mathrm{d}F_x.$$
This category was introduced by Lawvere in 1962, see here. Various papers mention as a clear fact that products do not exist in this category. Could someone please provide an argument that would make this fact clear also to me?
 A: We prove that a product of $(X,\mathbf{A})$ and $(Y,\mathbf{B})$ exists if and only if $\mathbf{A}=\{\emptyset,X\}$ or $\mathbf{B}=\{\emptyset,Y\}$.
Let $\mathcal{P}(X)$ denote the set of all probability measures on a measurable space $(X,\mathbf{A})$. For $x\in X$, let $\delta_x\in\mathcal{P}(X)$ be the point measure centered at $x$, that is, $\delta_x(A)=\mathbf{1}_A(x)$ for all $A\in\mathbf{A}$.
Given a morphism $F\colon(X,\mathbf{A})\to(Y,\mathbf{B})$ and a probability measure $p\in\mathcal{P}(\mathbf{A})$, denote $F(p)$ the probability measure $q\in\mathcal{P}(\mathbf{B})$ defined by $q(B)=\int_{x\in X}F_B(x)\mathrm{d}p$.
Lemma. Let $\mathbf{A}\neq\{\emptyset,X\}$ and $\mathbf{B}\neq\{\emptyset,Y\}$. Let $F\colon(Z,\mathbf{C})\to(X,\mathbf{A})$, $G\colon(Z,\mathbf{C})\to(Y,\mathbf{B})$ be probability kernels such that for all $p\in\mathcal{P}(X)$, $q\in\mathcal{P}(Y)$ there exists $r\in\mathcal{P}(Z)$ such that $F(r)=p$, $G(r)=q$. Then there exist distinct measures $s,s'\in\mathcal{P}(Z)$ such that $F(s)=F(s')$ and $G(s)=G(s')$.
Proof. Let $A_0\in\mathbf{A}\setminus\{\emptyset,X\}$, $B_0\in\mathbf{B}\setminus\{\emptyset,Y\}$. Let $x_0\in A_0$, $x_1\in X\setminus A_0$, $y_0\in B_0$, $y_1\in Y\setminus B_0$. For $i,j\in\{0,1\}$, let $r_{ij}\in\mathcal{P}(Z)$ be such that $F(r_{ij})=\delta_{x_i}$ and $G(r_{ij})=\delta_{y_j}$.
For $z\in Z$ denote $f(z)=F(z,X\setminus A_0)$, $g(z)=G(z,Y\setminus B_0)$. Then for all $i,j\in\{0,1\}$, $$\int f(z)\mathrm{d}r_{ij}=\delta_{x_i}(X\setminus A_0)=i,$$ $$\int g(z)\mathrm{d}r_{ij}=\delta_{y_j}(Y\setminus B_0)=j.$$
Denote $C_i=\{z\in Z\!:f(z)=i\}$ and $D_j=\{z\in Z\!:g(z)=j\}$. Then $r_{ij}(C_i)=r_{ij}(D_j)=1$. It follows that $r_{ij}(Z\setminus(C_i\cap D_j))=r_{ij}((Z\setminus C_j)\cup(Z\setminus D_j))\le r_{ij}(Z\setminus C_i)+r_{ij}(Z\setminus D_j)=0$, hence $r_{ij}(C_i\cap D_j)=1$.
For $u\in\big[0,\frac{1}{2}\big]$, let $s_u=u(r_{01}+r_{10})+\big(\frac{1}{2}-u\big)(r_{00}+r_{11})$. Then $s_u\in\mathcal{P}(Z)$. We have $s_u(C_0\cap D_1)=u$, hence $s_u\neq s_v$ whenever $u\neq v$. For all $A\in\mathbf{A}$ we have $$\int F_A(z)\mathrm{d}r_{ij}=F(r_{ij})(A)=\delta_{x_i}(A),$$ hence $$F(s_u)(A)=u\left(\int F_A(z)\mathrm{d}r_{01}+\int F_A(z)\mathrm{d}r_{10}\right)+\left(\frac{1}{2}-u\right)\left(\int F_A(z)\mathrm{d}r_{00}+\int F_A(z)\mathrm{d}r_{11}\right)=\frac{1}{2}\big(\delta_{x_0}(A)+\delta_{x_1}(A)\big).$$ It follows that $F(s_u)=\frac{1}{2}(\delta_{x_0}+\delta_{x_1})$, for all $u\in\big[0,\frac{1}{2}\big]$. Similarly, $G(s_u)=\frac{1}{2}(\delta_{y_0}+\delta_{y_1})$, for all $u\in\big[0,\frac{1}{2}\big]$. q.e.d.
Corollary. Let $(X,\mathbf{A})$, $(Y,\mathbf{B})$ be measurable spaces such that $\mathbf{A}\neq\{\emptyset,X\}$ and $\mathbf{B}\neq\{\emptyset,Y\}$. Then there does not exist a product of $(X,\mathbf{A})$ and $(Y,\mathbf{B})$ in the category of measurable spaces and probability kernels.
Proof. Assume that morphisms $F\colon(Z,\mathbf{C})\to(X,\mathbf{A})$, $G\colon(Z,\mathbf{C})\to(Y,\mathbf{B})$ form such a product. Let $W=\{w\}$. For any fixed measures $p\in\mathcal{P}(X)$, $q\in\mathcal{P}(Y)$ there exist morphisms $F'\colon\big(W,2^W\big)\to(X,\mathbf{A})$, $G'\colon\big(W,2^W\big)\to(Y,\mathbf{B})$ such that $F'_w=p$ and $G'_w=q$. By the universal property of a product, there exist a morphism $H\colon\big(W,2^W\big)\to(Z,\mathbf{C})$ such that $F'=F\circ H$ and $G'=G\circ H$. Denote $r=H_w$. Then $F(r)=p$ and $G(r)=q$, hence morphisms $F$, $G$ satisfy the assumptions of the Lemma. Hence there exist $s,s'\in\mathcal{P}(Z)$ such that $s\neq s'$, $F(s)=F(s')$ and $G(s)=G(s')$.
Let us take morphisms $K\colon\big(W,2^W\big)\to(Z,\mathbf{C})$, $K'\colon\big(W,2^W\big)\to(Z,\mathbf{C})$ such that $K_w=s$ and $K'_w=s'$. Then $K\neq K'$, $F\circ K=F\circ K'$ and $G\circ K=G\circ K'$, which contradicts the definition of a product in a category. q.e.d.
If $\mathbf{B}=\{\emptyset,Y\}$ then there exists a unique measure in $\mathcal{P}(Y)$, hence a unique morphism $G\colon(X,\mathbf{A})\to(Y,\mathbf{B})$. Let $F\colon(X,\mathbf{A})\to(X,\mathbf{A})$ be the identity morphism, that is, $F(x,A)=\mathbf{1}_A(x)$ for all $x\in X$, $A\in\mathbf{A}$. It is then easy to check that $F$ and $G$ form a product of $(X,\mathbf{A})$ and $(Y,\mathbf{B})$. Hence we obtain the following result.
Theorem. A product of $(X,\mathbf{A})$ and $(Y,\mathbf{B})$ in the category of measurable spaces and probability kernels exists if and only if $\mathbf{A}=\{\emptyset,X\}$ or $\mathbf{B}=\{\emptyset,Y\}$.
