Given an equivalence relation ${\sim} \in \Pi(U(X))$, is the meet of all the congruences $R \geq {\sim}$ a congruence? I'm trying to refine the notion of quotient objects in concrete categories, as it is presented in Theory of Mathematical Structures by Adámek. The motivation is how you can't always define a quotient of a group and an equivalence relation (or, equivalently, a subset of such group), but you can consider the normal closure of any subset of the group and then define the quotient group (which you associate to your initial equivalence relation, since the normal closure is unique). I'm also interested in presentations of algebraic structures by generators and relations.
What I'm trying to define as a quotient object $X/{\sim}$ in a concrete category $(C,U)$ is the quotient object $X/{\sim'}$, where $\sim'$ is the meet/infimum of all congruences which are greater or equal than $\sim$. Recall that an equivalence relation $R \in \Pi(U(X))$ is a congruence when the quotient $X/R$ exists.
However, for that definition to make sense, we need $\sim'$ to be a congruence! I considered the case where $C$ has products/coproducts and $U$ preserves products/coproducts, but I wasn't able to find the object $Y$ in $C$ such that $U(Y) \cong U(X)/{\sim'}$, which is the first step in the proof. We can assume $U$ is essentially surjective, which eliminates this first step, but now we need to construct a morphism $f \colon X \to Y$ such that $U(f)$ is the composition $U(X) \xrightarrow{\pi} U(X)/{\sim'} \xrightarrow{\cong} U(Y)$ (or, equivalently, $U(f) \cong \pi$), and we need more assumptions again.
What could some of those assumptions be? Is the property I'm trying to prove even true? I apologize, since I don't have a lot of knowledge in Category Theory.
[EDIT] I adapted the original definition found in the reference above in such a way that a quite exotic condition guarantees a stronger result. It is also "less evil" than its counterpart.

Definition:  Let $(C,U)$ be a concrete category, $X$ be an object of $C$ and $\sim$ an equivalence relation on $U(X)$. We say $Y$ is the quotient object of $X$ by $\sim$, which we denote by $X/{\sim}$, if the following conditions hold:

*

*$U(Y) \cong U(X)/{\sim}$ (in Adámek, equality is required);


*There exists a morphism $f \colon X \to Y$ such that $U(f) \cong \pi$ (in the arrow category), where $\pi$ is the projection $\pi \colon U(X) \to U(X)/{\sim}$;


*For every object $Z$ in $C$ and for every function $g \colon U(Y) \to U(Z)$ such that there exists a morphism $h \colon X \to Z$ satisfying $U(h) = g \circ U(f)$, there exists a morphism $\psi \colon Y \to Z$ such that $U(\psi) = g$.
If the quotient $X/{\sim}$ exists, we say $\sim$ is a congruence.

With that in mind, we have the following result:

Proposition: If the induced functor between arrow categories $\mbox{arr}(U) \colon \mbox{arr}(C) \to \mbox{arr}(\mbox{Set})$ is essentially surjective, then for every object $X$ in $C$ and every equivalence relation ${\sim} \in \Pi(U(X))$, it follows that $\sim$ is a congruence.

The classical example of this would be the category $\mbox{Top}$. To verify the condition, just consider the discrete topologies.
[EDIT 2] Here's a simple example to illustrate why just products/coproducts aren't really enough (to prove the meet is a congruence). Consider the cyclic group $C_8=(\{ 0,1, \dots, 7 \},+)$ and the equivalence relation $\sim$ on $\{ 0,1, \dots, 7 \}$ which correspondent partition is $\{ \{0,4\},\{1\}, \{2\}, \{3\}, \{5\}, \{6\}, \{7\} \}$. The congruences which are greater or equal than $\sim$ are:

*

*$\sim_1$, which corresponding partition is $\{ \{0,4\},\{1,5\},\{2,6\},\{3,7\} \}$;


*$\sim_2$, which corresponding partition is $\{ \{0,4,2,6\},\{1,5,3,7\} \}$;


*$\sim_3$, which corresponding partition is $\{ \{0,4,2,6,1,5,3,7\} \}$.
The meet of these congruences is $\sim_1$ (hence, according to our desired definition, the quotient $C_8/{\sim}$ is $C_8/{\sim_1}$). Also note that the quotients we have are $C_8/{\sim_1} \cong C_4, C_8/{\sim_2} \cong C_2, C_8/{\sim_3} \cong C_1$. However, the underlying set of the product $\prod_{i=1}^{3} C_8/{\sim_i} \cong \coprod_{i=1}^{3} C_8/{\sim_i}$ is not isomorphic to the underlying set of $C_8/{\sim_1}$.
 A: I guess that a good sufficient condition could be the following: $C$ closed for small limit and image and $U$ preserves these data.
$\newcommand{\cong}{\mathrm{Cong}_\sim}$
Assuming that $C$ satisfies the above mentioned condition you can consider the functor 
$$X/ \colon \cong (U(X)) \longrightarrow C$$
where $\cong(U(X))$ is the poset-category of congruence relations on the set $U(X)$ that contain $\sim$ and $X/$ associates


*

*to each congruence $\sim'$ the $C$-quotient $X/\sim'$

*to each inclusion $\sim'_1 \subseteq \sim'_2$ the unique morphism $$X/\sim'_1 \to X/\sim'_2$$ that makes commute the diagram with the quotient projections $X \to X/\sim'_i$ (where $i=1,2$).


This is a small diagram, because the class of congruence relations over the set $U(X)$ is a set, hence you can build the limit $\varprojlim_{\sim'} X/\sim'$.
This limit comes equipped with a natural morphism $$p \colon X \longrightarrow \varprojlim_{\sim'} X/\sim'$$ which is induced by the obvious cone over $X$, namely the cone formed by the quotient projections $X \to X/\sim'$.
If $C$ has images we can consider $\text{Im}(p)$, the image of $p$, with a little of set theoretic work one can show that $U(\text{Im}(p))$ is exactly $U(X)/\sim$, which proves that $\sim'$ is indeed a congruence.
I am not entirely sure if the result could be refined so that one can prove that actually $\varprojlim_{\sim'}X/\sim'$ is the lifting of the quotient $U(X)/\sim$, my guess would be that is not the case but I have no proof of it so I could be wrong on this.
Let me know if need additional details of something does not add up.
Edit: after a little bit I have come up with a different solution that does not reqire the existance of images.
Instead of requiring the existance of the image of $p \colon X \to \varprojlim_{\sim'} X/\sim'$ we can just require that $U(p) \colon U(X) \to U(\varprojlim_{\sim'}X/\sim')$ is a surjective function (i.e. an epimorphism of $\mathbf{Set}$).
When this condition is satisfied it is easy to see that the projection $U(p)$ factors through an isomorphism $U(X)/\sim \to U(\varprojlim_{\sim'}X/\sim')$, which proves that the quotient by $\sim$ lifts to a $C$-object.
On the other hand if $\bigwedge \cong(U(X))$ is a congruence then it is an element of $\cong (U(X))$ and $X/\bigwedge \cong(U(X))$ would be the "minimum" between the quotients considered.
Using this fact it should be easy to see that such $X/\bigwedge \cong(U(X))$ is a limit for the diagram $X/$.
So, assuming that $C$ has all small limit we have found that a necessary and sufficient condition for the existance of quotient object $X/\sim$ is the requirement that $$U(p) \colon U(X) \to U(\varprojlim_{\sim'} X/\sim')$$ is an epimorphism.
