In algebraic topology, there is a notion of covering space, which essentially "de-identifies" points that look the same but which for certain purposes really shouldn't be considered the same. I was wondering if we can do something like this in algebra. Explicitly, suppose we have an algebraic structure $Y$. Now imagine there is a subalgebra of $Y$ whose points we wish to "de-identify." We decribe this by another algebra $X$ and a homomorphism $\varphi : X \rightarrow Y$, and we think of the image of this homomorphism as being the subset of $X$ whose points we wish to de-identify so that $\varphi$ becomes an injection.
We're willing to tolerate further "de-identifications" than just those in the image of $\varphi$, but we want the minimum amount of de-identification.
General question. Is there a generally accepted way of "de-identifying" the elements of algebraic structures $X$ on the basis of a homomorphism $\varphi : X \rightarrow Y$ so as to make $\varphi$ injective?
Anyway, lets do the obvious thing. Firstly, define that an enlargement of the domain of $\varphi$ consists of an algebra $\overline{X}$ together with morphisms $X \hookrightarrow \overline{X}$ and $\overline{X} \rightarrow Y$ such that the former is injective and the obvious triangle commutes. Since enlargements of the domain of $\varphi$ form a category in a natural way, we can try looking for terminal objects in the category of enlargements.
Specific question. Let $\mathsf{T}$ denote a Lawvere theory and suppose that $\varphi : X \rightarrow Y$ denotes a morphism of $\mathsf{T}$-algebras. Does the category of enlargements of the domain of $\varphi$ always have a terminal object?
The answer to the specific question is no. For simplicity, we may as well look at the case where $Y$ is the terminal algebra. Then we can forget about $\phi$ and $Y$ entirely and just think about the category of enlargements of $X$. To further simplify matters, we could take $X$ to be initial algebra. Then the category of enlargements of $X$ is just the full subcategory of algebras $\bar{X}$ such that the unique homomorphism $X \to \bar{X}$ is injective.
Consider, for example, commutative rings. Then the category of enlargements of $\mathbb{Z}$ is the category of commutative rings of characteristic $0$. It is easy to see that this category has no terminal object: $\mathbb{Q}$ is in this category, so the terminal object would have to be a non-zero $\mathbb{Q}$-algebra $A$; but $\mathbb{Z} [x]$ is also in this category, and there is always more than one homomorphism $\mathbb{Z} [x] \to A$.