Does a would-be isomorphism between a known and suspected category object guarantee the latter object to be in the category?

Question

In this question from a few day ago, it became evident that, given a known vector space $(V,+,\times)$, a possible vector space $(W,\oplus,\otimes)$, and a transformation $T:V\rightarrow W$ that is is bijective and linear with respect to $+,\times,\oplus,\otimes$, then $(W,\oplus,\otimes)$ is indeed a vector space.

In other words, the existence of this would-be vector space isomorphism $T$ results in $(W,\oplus,\odot)$ being a vector space, since $W$ “should” be isomorphic with $V$.

Can this be generalized to other categories? That is, if $A$ is an object in a category, $B$ is suspected of being in that category, and $T:A\rightarrow B$ would be an isomorphism if we knew $B$ to be another object in that same category, does the existence of this $T$ guarantee $B$ to be another object in the same category as $A?$ If not, how general can this idea be beyond vector spaces?

If this is true, applications could be:

1. Finding a would-be isomorphism between a known and suspected vector space, rather than running through the vector space axioms for the suspected vector space (as in the link)

2. Finding a would-be homeomorphism between a known and suspected topological space, rather than showing the suspected open sets in the latter space to satisfy the axioms of being a topology

Answer 1

I think a useful general situation to ask about would be: suppose $\mathcal{C}'$ is a subcategory of $\mathcal{C}$. We could then define $\mathcal{C}'$ to be closed under isomorphisms if whenever $f \in \operatorname{Hom}_{\mathcal{C}}(X, Y)$ is an isomorphism, and $X \in \mathcal{C}'$, then $Y \in \mathcal{C}'$ and $f\in \mathcal{C}'$ also.

However, subcategories are not necessarily closed under isomorphisms. (Though "nicely defined" subcategories will tend to be closed under isomorphisms.) One possible example: if you take the category of vector spaces, then the full subcategory whose objects are $\{ \mathbb{R}^n \mid n \in \mathbb{N} \}$ would not be closed under isomorphisms, since for example the set of polynomials of degree $< n$ is isomorphic to $\mathbb{R}^n$ but is not literally equal to $\mathbb{R}^n$.

On the other hand, an example of a subcategory which would be closed under isomorphisms, and would generalize your example: suppose $\mathcal{C}$ is a variety of algebras, and $\mathcal{C}'$ is the full subcategory whose objects are the objects of $\mathcal{C}$ satisfying some additional set of identities (compared to the set of identities you already have in the definition of $\mathcal{C}$). Then $\mathcal{C}'$ is in fact closed under isomorphisms. Your example would be the case: $\mathcal{C}$ is the category of sets $V$ with a binary operation $+ : V\times V \to V$, a scalar multiplication (expressed as a family of unary operators $\lambda \cdot : V\to V$ for each $\lambda \in \mathbb{R}$), etc., but not necessarily satisfying the vector space axioms. Then the category of vector spaces is a subcategory of this type, so it is closed under isomorphisms.

(The topological space example doesn't quite fit under this class of examples. However, it is similar: the category of topological spaces is a full subcategory of the category of sets $X$ along with a subset $\mathcal{T}_X \subseteq \mathcal{P}X$, where morphisms are the functions $f:X\to Y$ such that whenever $V \in \mathcal{T}_Y$, we have $f^{-1}(V) \in \mathcal{T}_X$. Then this subcategory is indeed closed under isomorphisms.)

Answer 2

Let $$ be some field. Here is one way of looking at the given situation:

We have an auxiliary category $\mathscr{V}$ as follows.

• The objects of $\mathscr{V}$ are triples $(V, ⊕, ⊗)$ consisting of a set $V$ and two binary operations $$ ⊕ \colon V × V \to V \,, \quad ⊗ \colon × V \to V \,. $$ Note that we don’t require these operations to satisfy any further properties.
• A morphism from a $(V, ⊕, ⊗)$ to $(V', ⊕', ⊗')$ in $\mathscr{V}$ is a set-theoretic map $f$ from $V$ to $V'$ such that $$ f(v_1 ⊕ v_2) = f(v_1) ⊕' f(v_2) \,, \quad f(λ ⊗ v) = λ ⊗' f(v) $$ for all $λ ∈ $ and all $v, v_1, v_2 ∈ V$.
• The composition of morphisms in $\mathscr{V}$ is given by the usual composition of maps.

The category of $$-vector spaces, which we shall denote by $\mathbf{Vect}_$ is a full subcategory of $\mathscr{V}$. Your example relies on the following two observations:

• A morphism in $\mathscr{V}$ is an isomorphism if and only if it bijective.
• The category $\mathbf{Vect}_$ is closed under isomorphism inside of $\mathscr{V}$: if $V$ and $V'$ are two isomophic objects of $\mathscr{V}$ such that $V$ is contained in $\mathbf{Vect}_$, then $V'$ is also contained in $\mathbf{Vect}_$.

By using the auxiliary supercategory $\mathscr{V}$ of $\mathbf{Vect}_$ we can make sense of objects and morphisms “outside” of $\mathbf{Vect}_$.