Bijective correspondence between congruences on universal algebras I'm reading the chapter on univesal algebras in P. A. Grillet's Abstract Algebra. In Proposition 1.9 the author states:

Let $ \varphi\colon A\to B $ be a morphism of universal algebras. If $ \mathcal{E} $ is a congruence on $ B $, then $ \varphi^{-1}\mathcal{E} $ defined by
$$
{x\mathrel{\varphi^{-1}\mathcal{E}}y}\iff{\varphi x\mathrel{\mathcal{E}}\varphi y}
$$
is a congruence on $ A $. If $ \varphi $ is surjective, then $ A/\varphi^{-1}\mathcal{E}\cong B/\mathcal{E} $, and the above defines a one to one correspondence between the congruences on $ B $ and congruences on $ A $ that contain $ \operatorname{ker}\varphi $.

While I encounter no problem finding an isomorphism $ A/\varphi^{-1}\mathcal{E}\to B/\mathcal{E} $, the second part makes me a little bit unsure about how to proceed.
My attempt
Let $ \varphi $ be a surjective homomorphism $ A\to B $, and let's call $ \mathcal{E} $ a congruence on the universal algebra $ A $. Its image $ \varphi\mathcal{E} $ by the Cartesian product of $ \phi $ with itself
$$
\varphi\mathcal{E}=\left(\phi^2\right)_{*}\mathcal{E}
$$
is again a congruence on $ B $, since can easily find a homomorphism $ B\to A/\mathcal{E} $ such that $ \operatorname{ker}(\pi\upsilon)=\varphi\mathcal{E} $, by composing the canonical projection $ \pi\colon A\to A/\mathcal{E} $ and a right inverse $ \upsilon\colon B\to A $ of $ \varphi $
$$
\begin{array}{ccc}
B & \overset{\upsilon}{\longrightarrow} & A & \overset{\pi}{\longrightarrow} & A/\mathcal{E}
\end{array}
$$
(following the proof strategy that proving the first part of the claim involves). Let $ \upsilon $ be such a right inverse, and let $ x,y $ points of $ B $ such that $ \pi\upsilon(x)=\pi\upsilon(y) $ (i.e., let $ x\mathrel{\operatorname{ker}(\pi\upsilon)}y $); then $ \upsilon x\mathrel{\mathcal{E}}\upsilon y $, and thus $ x=\varphi\upsilon x $ is in relation with $ y=\varphi\upsilon y $. The converse should be quite the same. $ \square $
I'm here concerned about the right inverse $ \upsilon $ not being a homomorphism: it is not always the case.
My questions
Now, is the function
$$
\mathcal{F}\colon
\begin{align}
\left\{\text{congruences on $ A $ which contain $ \operatorname{ker}\phi $}\right\}&\to\left\{\text{congruences on $ B $}\right\}\\
\mathcal{E}&\mapsto \left(\varphi\times\varphi\right)_{*}\mathcal{E}
\end{align}
$$
by which each congruence $ \mathcal{E} $ is mapped in the set
$$
\left(\varphi\times\varphi\right)_{*}\mathcal{E}=\left\{\left(x,y\right):\text{$ x=\varphi x_0 $ and $ y=\varphi y_0 $ such that $ x_0\mathrel{\mathcal{E}}y_0 $}\right\}
$$
bijective? (I think so, but I have still to prove it; the inverse of $ \mathcal{F} $ must be the function that maps each congruence $ \mathcal{E} $ on $ B $ with the $ \phi^{-1}\mathcal{E} $ of the beginning of this post). Have I done the previous steps correctly? And, most important, is this kind of "correspondence" that the author had in mind?
I have another question: in what manner the isomorphism $ A/\varphi^{-1}\mathcal{E}\cong B/\mathcal{E} $ is related to this proof? (Maybe it isn't, and the claim of the existence of $ \mathcal{F} $ is only inserted in the same proposition with no direct relation with such, but...)
 A: Let $\varphi:\mathbf A \to \mathbf B$ be a homomorphism.
For each congruence $\beta$ of $\mathbf B$, define the binary relation $\alpha$ on $A$ by making
$$a \,\alpha\, b \;\Longleftrightarrow\; \varphi(a) \,\beta\, \varphi(b).$$
Apparently you have no difficulty in recognizing this defines a congruence $\alpha$ on $\mathbf A$ (where $\alpha = \varphi^{-1}\beta$).
Now suppose $\varphi$ is surjective.
Let $\beta$ and $\beta'$ be congruences on $\mathbf B$ and $\alpha, \alpha'$ be congruences on $\mathbf A$ such that $\alpha = \varphi^{-1}\beta$ and $\alpha' = \varphi^{-1}\beta'$.
To prove the correspondence is one-to-one (that is, injective) we must show that if $\alpha=\alpha'$ then $\beta=\beta'$.
So let $\alpha = \alpha'$ and take $u,v \in B$ such that $(u,v) \in \beta$.
Since $\varphi$ is onto, there exist $a,b \in A$ such that $\varphi(a)=u$ and $\varphi(b)=v$.
Now,
\begin{align}
(u,v) \in \beta
&\;\Longleftrightarrow\; (\varphi(a),\varphi(b)) \in \beta\\
&\;\Longleftrightarrow\; (a,b) \in \alpha\\
&\;\Longleftrightarrow\; (a,b) \in \alpha'\\
&\;\Longleftrightarrow\; (\varphi(a),\varphi(b)) \in \beta'\\
&\;\Longleftrightarrow\; (u,v) \in \beta'.
\end{align}
Hence $\beta = \beta'$ and the correspondence is one-to-one.
Now let, again, $\beta$ be a congruence on $\mathbf B$ and $\alpha$ be the congruence on $\mathbf A$ defined by $\alpha = \varphi^{-1}\beta$.
Let $a,b \in A$.
\begin{align}
(a,b) \in \ker\varphi
&\;\Longrightarrow\; \varphi(a) = \varphi(b)\\
&\;\Longrightarrow\; \varphi(a) \,\beta\, \varphi(b)\\
&\;\Longrightarrow\; a \,\alpha\, b,
\end{align}
where the second implication follows from the reflexivity of $\beta$.
Thus $\ker\varphi \subseteq \varphi^{-1}\beta$.
