This is a question which has been bothering me a lot since I've began studying abstract mathematics. I'm not sure if others are also bothered by this, or if this is something not to worry about.
A lot of times, there is a statement about extending an object to a larger one, such that the first is a sub-object of the other. And a lot of places, we replace the object by an isomorphic (in some sense) copy of itself, which is not really the question, because we wanted to create a larger object containing the first as a subset.
For example, field extensions: if $F$ is a field and $p(x)$ is an irreducible polynomial over this field, then there is a field extension $K/F$ such that $p(x)$ has a root in $K$. A usual proof of this is to consider $F[x]/(p(x))$, where we identify $F$ with a copy of it in $F[x]/(p(x))$, which more or less gets the job done, but I'm still not able to convince myself that there is a field $K$ with the given property such that $F\subset K$.
There are many more examples of situations like this (I am not able to recall a lot of them), where people just work with isomorphic copies.
So my question is: are mathematicians fine to just work with isomorphic copies, and how in particular do I convince myself in the above question?
This answer addresses the question of how to prove an extension exists which contains $F$ set-theoretically as a subfield.
Let $F'$ be the isomorphic image of $F$ sitting inside $K' = F[x]/(p(x))$ consisting of constants. Naïvely, one would like to simply replace the elements of $F'$ with elements of $F$. The problem is that it is not at all clear that there can't be elements of $K' - F'$ that also belong to $F$.
There are a few possible solutions I see to this objection.
First solution (using the axiom of foundation). Use the set-theoretical axiom of foundation to show that this doesn't occur. Namely, say $f \in F$ also occurs in $K'$.
An element of $K'$ is an equivalence class $A$ in $F[x]$. But since $p(x)$ is nonzero, a representative $q(x) \in A$ can have any leading coefficient whatsoever, including $f$. (If $f = 0$, we consider instead the zero coefficient above the leading coefficient of $q$.) Now the polynomial $q(x)$ is set-theoretically a sequence of elements in $F$ one of whose members is $f$. Working through the definitions--things may vary slightly depending on your conventions--we get a sequence of the form $$f \in u_1 \in u_2 \in \dots \in u_r \in q \in A,$$ where $r$ is a small integer that can be determined explicitly. This contradicts the axiom of foundation since we're assuming $A = f$.
This solution is perhaps not easily generalizable to other situations, so a different approach is preferable.
Second solution (preferred). For each element of $K' - F'$ that also occurs as an element $f \in F$, replace that copy of $f$ with the pair $(f,\alpha)$, where $\alpha$ is the smallest ordinal such that $(f, \alpha) \not\in K' \cup F$. Such an ordinal exists, because otherwise there would be an injection of the proper class of all ordinals into the set $K' \cup F$.
Third solution (with the axiom of choice). It is enough to prove the existence of some set $A$ that has the same cardinality as $K' - F'$ which is disjoint from $F$. Then the field structure on $K'$ can be transported onto $A \cup F$ in the obvious way. Of course, the second solution accomplishes this, but it's really easy to see using cardinal arithmetic.
Namely, pick some infinite set $S$ that is strictly larger than both $K'$ and $F$. (For example, we could take $S = P(K' \cup F) \cup \mathbf{N}$.) Then using the axiom of choice, $S - F$ has the same cardinality as $S$, which itself contains a set equipotent to $K' - F'$.
This relies on the fact that if $B$ and $C$ are sets with cardinality $< \kappa$ (with $\kappa$ infinite), then $B \cup C$ also has cardinality $< \kappa$.