Say $F$ is a functor from category $C$ to $D$. By "fully faithful", I mean $f \mapsto Ff$ is injective ("faithful") and surjective ("full") in $C(X, Y) \to D(FX, FY)$.

My question is: when $F$ is fully faithful, why does $FX \cong FY$ imply that $X \cong Y$ for objects $X, Y$ in $C$?

The best I am able to come up with is:

Since $FX \cong FY$, there exists an isomorphism $h \in D(FX, FY)$. And, because $F$ is full, there exists a morphism $f \in C(X, Y)$ such that $F f = h$, and also a morphism $g \in C(Y, X)$ such that $F g$ is the inverse of $h$.

but... I don't know how to prove that $f \in C(X, Y)$ is an isomorphism. I think I need to show that $f g = \text{id}_Y$ and $g f = \text{id}_X$, but I don't know how.

Notation background: I'm using the text

Bradley, T. D., Bryson, T., & Terilla, J. (2020). Topology: A Categorical Approach. MIT Press.

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    $\begingroup$ Your question asks "My question is: when $F$ is fully faithful, why does $FX \cong FY$ imply that $X \cong Y$?" This has been answered; I just want to note that "conservative" or "reflects isomorphisms" does not mean $FX \cong FY$ implies $X \cong Y$ - at least not according to the usual definitions! "Conservative" means "if $Ff$ is an isomorphism then $f$ was an isomorphism: en.wikipedia.org/wiki/Conservative_functor. You are talking about some other property, which neither implies nor is implied by conservativity. $\endgroup$
    – John Baez
    May 11, 2023 at 20:22
  • $\begingroup$ @JohnBaez I don’t know whether this terminology is standard but the nLab calls functors $F$ for which $F(X)\cong F(Y)$ implies $X\cong Y$ essentially injective. Full conservative functors are of course essentially injective. $\endgroup$ Sep 20, 2023 at 7:55

2 Answers 2


You're nearly there! You've used the fact that $F$ is full, now you need to use the fact that it's faithful...

Suppose $f\in C(X,Y)$ and $g\in C(Y,X)$, such that $Ff\in D(FX,FY)$ and $Fg\in D(FY,FX)$ are inverses. Then $F(f\circ g) = Ff\circ Fg = \text{id}_{FY} = F(\text{id}_Y)$, and since $F$ is faithful, $f\circ g = \text{id}_Y$. The same argument shows $g\circ f = \text{id}_X$.

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    $\begingroup$ Beat me to it. Hi Alex! $\endgroup$ Jan 26, 2022 at 18:08
  • $\begingroup$ @AlfredYerger Hi Tom $\endgroup$ Jan 26, 2022 at 18:46
  • $\begingroup$ thank you so much! follow up question - why can i assume $Fg$ is both the left and right inverse of $Ff$? i thought i'd maybe have to say, "because $Ff$ is an isomorphism, it has left and right inverses, and because $F$ is full, there exist morphisms $f^{-1}_l, f^{-1}_r \in C(Y, X)$ such that $F f^{-1}_l$ and $F f^{-1}_r$ are the left and right inverses, respectively, of $F f$." And then proceed from there to show $f^{-1}_l$ and $f^{-1}_l$ are the left and right inverses, respectively, of $f$, proving its an isomorphism. Why do I not have to worry about this? $\endgroup$ Jan 26, 2022 at 21:50
  • $\begingroup$ maybe it was a dumb question. are left and right inverses always the same for isomorphisms? $\endgroup$ Jan 26, 2022 at 21:59
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    $\begingroup$ @grisaitis (1) "$f$ is an isomorphism" means there is an arrow $g$ such that $fg = \mathrm{id}$ and $gf = \mathrm{id}$. (2) Suppose $f$ is an arrow (not necessarily an isomorphism) which has a left inverse $g$ (so $gf = \mathrm{id}$) and also a right inverse $h$ (so $fh = \mathrm{id}$). Then $g = g\mathrm{id} = gfh = \mathrm{id}h = h$. So $f$ is an isomorphism after all. $\endgroup$ Jan 26, 2022 at 22:17

A necessary and sufficient condition for a full functor $F\colon C\to D$ to be conservative is that $Fg\colon FY\to FY$ being the identity implies $g\colon Y\to Y$ is an isomorphism (e.g. that there exists $f$ and $h$ such that $fg$ and $gh$ are identities, as then $f=fgh=h$). Since a faithful functor has $Fg\colon FY\to FY$ the identity morphism only if $g\colon Y\to Y$ is already the identity morphism, it follows that full and faithful functors are conservative.

The necessity of the condition is immediate because $F$ being conservative implies that if $Fg$ is the identity morphism, which is an isomorphism, then $g$ is an isomorphism.

For sufficiency: if $Fg\colon FX\to FY$ is an isomorphism, i.e. if there are $\phi\colon FY\to FX$ and $\psi\colon FX\to FY$ such that $\phi Fg$ and $Fg\psi$ are identites, then fullness implies there exists $f\colon Y\to X$ and $h\colon X\to Y$ with $Ff=\phi$ and $Fh=\psi$, which then have the property that $FfFg=F(fg)$ and $FgFh=F(gh)$ are identities. Then the desired property implies $fg\colon X\to X$ and $gh\colon Y\to Y$ are isomorphisms, whence $((fg)^{-1}f)g$ and $g(h(gh)^{-1})$ are identities, so $g$ is an isomorphism, as desired.


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