# Recommendation on setting the reference axis for mathematical objects

(I don't know what the title should be for this post, please change it if you have a better title. Also tags)

In many situations, there arises cases that one mathematical structure embeds into another mathematical structure. Moreover, some structures are defined as these relations, namely universal mapping property.

For example, the standard construction of a completion $M$ of a metic space $X$ (i.e. via Equivalence classes of cauchy sequences) is a set NOT containing $X$. That is, $X\not\subset M$. However, there is a trick to transform this to a set $S$ which has identical mathematical meaning of $M$ and containing $X$.

To be clear, here is a trick to transform a group $G$ to which an algebraic structure $M$ embeds, to indeed a group $H$ literally containing $M$.

Yes, this trick is extremely messy.

However, I found this trick necessary if one wants to define "the reference" object.

For example, via this trick, one can define "the completion $M$ of a metric space $X$ such that $X\subset M$" and "the field of quotients $F$ of an integral domain $R$ such that $R\subset F$ and "the free group $F(S)$ on a set $S$ such that $S\subset F(S)$ and etc.

When I first learned this trick, I felt happy and satisfied with the fact that I can make an output object to be a set containing a input object.

However, the more I define mathematical objects in this way, I feel like I'm not doing mathematics but working in a factory, since this trick is extremely artificial.

However, if one does not define the reference axis, there should always be one more line in proofs, that is, "Let $f$ be the canonical map".

I really want to know what people think on my opinion. I think this process is foolish but it deserves. What do you think?

What one really does when completing a metric space $X$, is to find a complete metric space $\overline{X}$ together with an isometric map $i : X \to \overline{X}$. This map should belong to the definition of a completion (unfortunately, many authors forget this). It has the universal property that for every complete metric space $Y$ and every isometric map $f : X \to Y$, there is a unique isometric map $\overline{f} : \overline{X} \to Y$ such that $\overline{f} \circ i = f$. (This is best remembered via some commutative diagram, draw it!) The fact that $i$ is injective allows us, in principle, to identify the underlying set of $X$ with the underlying set of $\overline{X}$, but this is irrelevant when working with these objects as soon as we remember our $i$. In some sense, $i$ witnesses this kind of identification, and it is good to keep track of it.
Similarly, if $R$ is an integral domain with field of fractions $Q(R)$, this comes equipped with an injective homomorphism $i : R \to Q(R)$, $i(r)=\frac{r}{1}$. If $f : R \to K$ is any injective homomorphism into a field $K$, then it induces a unique homomorphism $\overline{f} : Q(R) \to K$ such that $\overline{f} \circ i = f$. I would not recommend to identify $r$ with $i(r)$. Don't confuse $i$ with the identity, it is not the identity. This becomes even clearer when you look at the more general construction of localization. Here, one starts with an arbitrary commutative ring $R$ and a multiplicative set of elements $S \subseteq R$ which one would like to invert formally. This is done via some ring $S^{-1} R = \{\frac{r}{s} : r \in R, s \in S\}$ equipped with a homomorphism $i : R \to S^{-1} R$. But $i$ is not injective when $S$ contains zero divisors! In fact, if $rs=0$ with $r \neq 0$, we have $\frac{r}{1}=\frac{rs}{s}=0$, but $r \neq 0$.
Another example comes from field theory. If $K$ is a field and $f \in K[x]$ is an irreducible polynomial, one would like to adjoin a formal root $\alpha$ of $f$ to $K$. This is usually done by constructing $K[x]/(f)$. Notice again that $K$ is not really a subset of $K[x]/(f)$, but that we have a field homomorphism $K \to K[x]/(f)$. This is what matters. For example, $\mathbb{R}$ extends to $\mathbb{C} := \mathbb{R}[x]/(x^2+1)$. Even though $\mathbb{R}$ is not a subset of $\mathbb{C}$ this way, this doesn't matter at all, since we only need a field homomorphism $\mathbb{R} \to \mathbb{C}$. In my opinion, the correct definition of a field extension of $K$ is just: It is a homomorphism of fields $i : K \to L$. We may also call $L$ an extension field of $K$ via $i$. For example, $\mathbb{R}(x,y)$ is an extension field of $\mathbb{R}(t)$ in many ways, two examples are $t \mapsto x$ and $t \mapsto y$. There are many homomorphisms $\mathbb{R}(t) \to \mathbb{R}(x,y)$ and no one is preferred to the other.
More generally, if $A,B$ are (algebraic) structures of the same type, we should not think of $A$ as a subobject of $B$ (or $B$ an extension of $A$) when the underlying set of $A$ is a subset of the underlying set of $B$ and the operations are compatible, but rather when we are given an injective homomorphism (or monomorphism) $i : A \to B$. In this definition, being a subobject is not really a relation between two objects. Rather, the homomorphism $i : A \to B$ witnesses the subobject relation and therefore should be seen as the subobject. For more on this, see nlab/subobjects, math.SE/295800, math.SE/704593.