I am trying to prove the following:
Let $A$ and $B$ be sets endowed with the equivalence relations $\sim_A$ and $\sim_B$, respectfully. Define the relation $\sim$ on $A \times B$ by setting
$(a_1,b_1) \sim (a_2,b_2) \; \text{if and only if} \; a_1\sim_A a_2 \; \text{and} \; b_1\sim_B b_2.$
Use the universal property of quotients to establish that there are functions $(A\times B)/\sim \rightarrow A/\sim_A$, $(A\times B)/\sim\rightarrow B/\sim_B$.
Prove that $(A\times B)/\sim$ with these two functions satisfies the universal property for the product of $(A/\sim_A)\times (B/\sim_B)$.
Conclude without further work that $(A\times B)/\sim \cong (A\sim_A)\times (B/\sim_B)$.
First of all, sorry that I cannot draw the diagrams here. Hopefully, this isn't too terrible.
I worked out the first two parts without much difficulty: let $\varphi_A$ and $\varphi_B$ denote the maps which sends $(a,b)$ to $[a]_{\sim_A}$ and $(a,b)$ to $[b]_{\sim_B}$, respectively. Then, by the universal property of quotient spaces, there exists unique functions $\psi_A:(A\times B)/\sim \rightarrow A/\sim_A$ and $\psi_B:(A\times B)/\sim \rightarrow B/\sim_B$ such that $\varphi_A=\psi_A\circ \pi$ and $\varphi_B=\psi_B\circ \pi$, where $\pi:A\times B\rightarrow (A\times B)/\sim$ is the projection map. Thus, by the universal property for $(A/\sim_A)\times (B\sim_B)$ we can conclude there exists a unique function $\Psi:A\times B\sim\rightarrow (A/\sim_A)\times(B/\sim_B)$ such that $\pi_A\circ \Psi=\psi_A$ and $\pi_B\circ \Psi=\psi_B$, where $\pi_i$ represents the projection onto the $i$ coordinate. Namely we get $\Psi=\psi_A\times\psi_B$, where the uniqueness of $\Psi$ follows from the uniqueness of $\psi_A$ and $\psi_B$.
From here I don't see how to conclude this is an isomorphism. Any ideas? Have I made a mistake?