Topology: Product of quotient spaces vs. quotient of product spaces

Question

For an equivalence relation $\sim$ on a topological space $X$, form the natural $f^{\ast}$ map from the quotient space $(X \times X)/\approx$ to $(X/\sim) \times (X/\sim)$, where $\approx$ is the equivalence relation defined coordinatewise on the product $X \times X$, that is, $ (x, y) \approx (u, v)$ if and only if $x \sim u$ and $y \sim v$. In other words, $f^{\ast}$ is the map obtained from the map $f = q \times q : X \times X \rightarrow (X/\sim) \times (X/\sim)$ by passing to the quotient, where $q : X \to X/\sim$ is the quotient map associated with $\sim$.

Thus $f^{\ast}$ is the unique continuous map such that $f^{\ast} \circ \pi =q \times q = f$, where $\pi \colon X \times X \rightarrow (X \times X)/\approx$ is the quotient map. Explicitly, for $(x, y) \in X \times X$, we have $f^{\ast}([x, y]_{\approx}) = (q(x), q(y)) = ([x]_{\sim}, [y]_{\sim})$, where $[.\, ,.]_{\approx}$ denotes the equivalence class in $X \times X$ for $\approx$ and $[.]_{\sim}$ denotes the equivalence class in $X$ for $\sim$.

It is well known that $f^{\ast}$ is a bijective continuous map, that $f^{\ast}$ need not be a homeomorphism, and that in some circumstances (by a theorem of Whitehead) $f^{\ast}$ is a homeomorphism.

1. What is the situation when $X = \mathbb{Q}$, the space of rationals, and $\sim$ collapses the set $\mathbb{Z}$ of integers to a point — that is, $x \sim y$ if and only if $x = y$ or else $x$ and $y$ are both integers? Is $f^{\ast}$ a homeomorhism here?
2. What is the situation when $X = \mathbb{R}$, the space of reals, and $\sim$ collapses the set $\mathbb{Q}$ of rationals to a point — that is, $x \sim y$ if and only if $x = y$ or else $x$ and $y$ are both rational? Is $f^{\ast}$ a homeomorhism here?

In general, if $f = q \times q$ fails to be a quotient map, then $f^{\ast}$ cannot be a homeomorphism, because a homeomorphism is a quotient map and then the composite $f^{\ast} \circ \pi$ of two quotient maps would again be a quotient map. The conoverse is true as well. So in each case, the problem amounts to asking whether $f = q \times q$ is also a quotient map.

Related:

1. When is the product of two quotient maps a quotient map? . A comment to that Q's answer https://math.stackexchange.com/a/31700/32337 references an example in Ronnie Brown's book "Topology and Groupoids, namely, that the product map $q \times 1_{\mathbb{Q}} \colon \mathbb{Q} \times \mathbb{Q} \rightarrow \mathbb{Q}//\mathbb{Z} \times \mathbb{Q}$ is not a quotient map, where $\mathbb{Q}//\mathbb{Z}$ is obtained by collapsing the set of integeer to a pooint and $q$ is the quotient map. (Significant detail of the proof is left to the reader in Brown's book!)
2. Product of quotient topology not homeomorphic to the quotient of product topology . The answer https://math.stackexchange.com/a/4920555/32337 references Example 7, Section 22, of Munkres Topology, namely that the product $q \times 1_{\mathbb{Q}} \colon \mathbb{R} \times \mathbb{Q} \rightarrow \mathbb{R}//\mathbb{Z}_{+}$ is not a quotient map, where $\mathbb{R}//\mathbb{Z}_{+}$ is the quotient of $\mathbb{R}$ obtained by collapsing the set $\mathbb{Z}_{+}$ to a point and $q$ is the quotient map.
3. Exercise 6 in Section 22 of Munkres Topology gives an example of the same kind as the first one I ask about except that the quotient space in his example is not a Hausdorff space: the product map $q \times q \colon \mathbb{R}_K \times \mathbb{R}_K \rightarrow (\mathbb{R}_K//K) \times (\mathbb{R}_K//K)$ is not a quotient map, where $\mathbb{R}_K$ is the reals with its "$K$-topology", $\mathbb{R}_K//K$ is the quotient space obtained by collapsing the set $K = \{1/n \colon n = 1, 2, \dots\}$ to a point, and $q$ is the quotient map.

Answer 1

This answer can contain a gap. I am going to revisit it and to ensure that $\pi(U)$ is open. This can be already true but need to be checked or in can be true if we replace the Euclidean distance in the definition of $U$ by the $\ell_\infty$ distance.

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Let $Y$ is $\mathbb Z$ in the first case and $\mathbb Q$ in the second case. As I understand the definition of $f^*$, in order to show that it is not open (at $[Y\times Y]$), it suffices to find a neighborhood $U$ of $Y\times Y$ in $X\times X$ such that there exist no neighborhood $V$ of $Y$ with $V\times V\subset U$. For this choose any enumeration $\{y_n\}_{n\in\omega}$ of $Y\times Y$ and put $$U=\bigcup_{n\in\omega} \{x\in X\times X: d(x,y_n)<2^{-n}\},$$ where $d$ is the Euclidean distance on $X\times X$.