# Why are fully faithful functors conservative? (Or, why are isomorphisms reflected?)

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.

• 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. Commented May 11, 2023 at 20:22
• @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. Commented Sep 20, 2023 at 7:55

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 $$g\circ f = \text{id}_X$$.

• Beat me to it. Hi Alex! Commented Jan 26, 2022 at 18:08
• @AlfredYerger Hi Tom Commented Jan 26, 2022 at 18:46
• 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? Commented Jan 26, 2022 at 21:50
• maybe it was a dumb question. are left and right inverses always the same for isomorphisms? Commented Jan 26, 2022 at 21:59
• @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. Commented 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.