If $F: \mathscr{A} \rightarrow \mathscr{B}$ is an equivalence of categories, and $A,\tilde{A} \in \mathrm{obj}(\mathscr{A})$ such that $\tilde{A} \subseteq A$ in $\mathscr{A}$, does it immediately hold that $F(\tilde{A}) \subseteq F(A)$ in the category $\mathscr{B}$?
Just trying to sanity check here! If I think of the example of certain representations/invariant vector spaces which may be thought of as modules, this seems clear, but I'm struggling to explicitly show this! Though I am tempted to say that this property is not sufficiently `categorical' in the sense of answers such as: Properties preserved under equivalence of categories
Best,
M
$\newcommand{\A}{\mathscr{A}}\newcommand{\B}{\mathscr{B}}$This is true.
Say $h:a\to a’$ is a monic in $\A$ (the standard meaning of $a\subseteq a’$ is that such an $h$ exists); we want to show that $F(h)$ is monic. Let $\eta:GF\cong1_{\A}$ be any natural isomorphism; at least one exists since $F,G$ are pseudoinverse.
Suppose there is a $b\in\B$ and $f,g:b\to F(a)$ with $F(h)\circ f=F(h)\circ g$.
Then: $$\eta_{a’}GF(h)\circ G(f)=\eta_{a’}GF(h)\circ G(g)$$So: $$h\eta_a\circ G(f)=h\eta_a\circ G(g)$$So (since $h\eta_a$ is monic as a composite of monics) we have $G(f)=G(g)$. Because $G$, as an equivalence, must be faithful, we find $f=g$ as desired. Thus $F(h)$ is monic.
Another way to see this: let $h:a\to a’$ be a monic. This is exactly the same thing as saying that the arrows: $$h^\ast:\A(a’’,a)\to\A(a’’,a’)$$Are monic for all $a’’\in\A$ (that is, they are injective).
Fix $b\in\B$. Since $F$ is essentially surjective, there exists some $a’’\in\A$ with $F(a’’)\cong b$. Therefore there is a commutative diagram: $$\require{AMScd}\begin{CD}\B(b,F(a))@>F(h)^\ast>>\B(b,F(a’))\\@V\cong VV@VV\cong V\\\B(F(a’’),F(a’))@>F(h)^\ast>>\B(F(a’’),F(a))\\@V F\,\,\cong VV@VV\cong\,\,FV\\\A(a’’,a)@>>h^\ast>\A(a’’,a’)\end{CD}$$
Where the vertical edges are all isomorphisms. Therefore the top arrow is monic if and only if the bottom one is; $F(h)$ is monic if and only if $h$ is.