Let $X_{i\le n}$ be compact but not necessarily $T_2$, each with equivalence relation $\sim_k\ $.
Let $X:=\prod X_i, Y:=\prod\big(\ ^{X_k}/_{\sim_k}\big)\ $. Is it possible to find equivalence relation $\sim$ such that $X/\ _{\sim}\simeq Y ?$
Disproof:
We show first that possible equivalence relation for it must be unique,
1.$\ $Construct $\sim$ on $X$ such that $x\sim y$ iff $x_k\sim_k y_k$ for each coordinate. There obviously exists bijection between sets $X/_\sim $ and $Y$.
2.$\ $For uniqueness, if there exist some $\sim'$ on $X$ and some homeomorphism $h$ between $X/_{\sim'}$ and $Y$, then this induces quotient map from $X\to X/_{\sim'}\ \ $, constant exactly on product of each equivalence class of $\sim_k$, hence $\ \sim'=\sim\ .$ (up to homeomorphism)
Now we just need to prove that product of these quotient maps is not a quotient map.
$\textbf{3.}\ \ $Let $\mathbb R^*$, $\mathbb Q^*$ be one point compactifications of $\mathbb R$ and $\mathbb Q$ respectively.
Consider $\mathbb R^*/\mathbb Z$ obtained by identiying all integers to a point, by quotient map $p$. By assuming the truth of the problem, $p\times id_{\mathbb Q^*}$ is quotient map. We also have restriction of three maps to respective sets are quotient maps, denoted $q, i, q\times i$.
$\textbf{4.}\ \ $The map $q$ is quotient map and with the quotient map from $\mathbb R\to \mathbb R/\mathbb Z$ (by identifying all integers to a point), obviously induces homeomorphism between $\mathbb R/\mathbb Z\ $ and $p(\ \mathbb R^*-\{\infty\}) \ .$ But it is known that $q\times i$ cannot be a quotient map. Contradiction. (see https://i.sstatic.net/0IAjI.png)
Best Answer
By expanding every details, I get the following:
Disproof:
We show first that possible equivalence relation for it must be unique,
1.$\ $Construct $\sim$ on $X$ such that $x\sim y$ are equivalent iff $x_k\sim_k y_k$ for each coordinate. There obviously exists bijection between both set of equivalence classes by mapping product of equivalence classes of each $\sim_i$ to equivalence class of $\sim$.
2.$\ $For uniqueness (up to homeomorphism), if there exist some $\sim'$ on $X$ and some homeomorphism between $X/_{\sim'}\ \ $ and $Y$. Consider the quotient map from $X\to X/_{\sim'}\ \ $, constant exactly on product of each equivalence class of $\sim_k$, hence on each equivalence class of $\sim$. Now by considering the quotient space $X/_{\sim}\ \ $ and the respective quotient map from $X\to X/_\sim\ \ $, obviously induces homeomorphism between $X/_\sim$ and $X/_{\sim'}\ \ $, hence this resolves to homeomorphism between $X/_\sim$ and $\prod\bigg( \ ^{X_i}/_{\sim_i}\bigg)\ \ $.
Now we just need to prove that product of these quotient maps is not a quotient map.
e.g. Let $\mathbb R^*$, $\mathbb Q^*$ be one point compactifications of $\mathbb R$ and $\mathbb Q$ respectively.
Consider $\mathbb R^*/\mathbb Z$ obtained by identiying all integers to a point, by quotient map $p$. By assuming the truth of the problem, $p\times id_{\mathbb Q^*}$ is quotient map. Since these compactifications are $T_1$, hence $\mathbb R, \mathbb Q$ are open saturated in respective space. We also have restriction of three maps to respective sets are quotient maps, denoted $q, i, q\times i$.
The map $q$ with the quotient map from $\mathbb R\to \mathbb R/\mathbb Z$ (by identifying all integers to a point), obviously induces homeomorphism between $\mathbb R/\mathbb Z\ $ and $p(\ \mathbb R) \ .$ Next, since $id_{\mathbb Q^*}$ is a homeomorphism, we have its restiction $i$ is also a homeomorphism hence $i(\mathbb Q)$ has the standard topology $\mathbb Q$. By composing with suitable homeomorphism, we can modify the codomain such that the map $q\times i$ become a quotient map between $\mathbb R\times \mathbb Q$ and $\mathbb R/\mathbb Z\times \mathbb Q$. But it is known that $q\times i$ cannot be a quotient map. Contradiction. $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $(see https://i.sstatic.net/0IAjI.png)