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Is the following a true statement?

Let $F:\bf{A}\to\bf{B}$ be a functor, and let $\bf{C}$ be a full subcategory of $\bf{B}$ with inclusion functor $I:\bf{C}\to\bf{B}$. If $G:\bf{C}\to {A}$ is a functor such that $F\circ G\cong I$, and $\mathscr{B}=F(\mathscr{A})$ for some object $\mathscr{A}$ of $A$, then $G(\mathscr{B})\cong \mathscr{A}$.

Any help is appreciated.

EDIT: replaced $F(\mathscr{B})\cong \mathscr{A}$ with $G(\mathscr{B})\cong \mathscr{A}$

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    $\begingroup$ The question does not make any sense as written; $b = F(a)$ is an object of $B$ so we can't apply $F$ to it. If the intended question was $G(b) \cong a$, there is already no reason for this to be true when $A, B, C$ are discrete categories (so, sets); take $G$ to be the inclusion of a point and $a$ to be anything outside the image of $G$. What is your motivation? $\endgroup$
    – Qiaochu Yuan
    Commented Nov 9, 2017 at 1:06
  • $\begingroup$ Yes, I did mean $G(\mathscr{B})\cong\mathscr{A}$. Thank you for the counterexample, it is much appreciated. $\endgroup$
    – clanijos
    Commented Nov 9, 2017 at 1:14
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    $\begingroup$ Hochster in his paper "Prime ideal structure in commutative rings", proves that Spec(_) is 'invertible' in the sense I've stated in the original post on the full subcategory of the category of spectral spaces consisting of $T_1$ spectral spaces. I'm interested in recovering rings from objects in a full subcategory of that subcategory. $\endgroup$
    – clanijos
    Commented Nov 9, 2017 at 1:21
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    $\begingroup$ It would make much more sense then to state this specific instance in the question and then ask whether it can be made abstract nonsense, rather than starting with abstract nonsense from which the original motivation cannot be guessed at all. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Nov 9, 2017 at 6:17

1 Answer 1

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[No need to use such baroque fonts. Fancier fonts usually means fancier mathematical object, but you are going the other way around: plain font = category, fancy font = object of category. I will use lower-case letters for objects of categories, represented by plain upper-case letters]

I'm not convinced your question makes sense yet, even after the correction. How on earth can you apply $G$ to the object $b$ of $B$ unless $b$ is already an object of $C$? If you assume $b$ is really an object of $C$ (that is, is in the image of $I$, or at least iso to something in that image, but will assume the easier case), then you have $F(a) = b\simeq F(G(b))$ so if $F$ is full on isomorphisms (or "creates isomorphisms"), you're done, since you can lift $F(a) \simeq F(G(b))$ to an isomorphism $a \simeq G(b)$.

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