Functors $F\colon C\to D$ preserve diagrams: if a diagram $A$ commutes in $C$, then $FA$ commutes in $D$ (where $FA$ is the diagram in $D$ that can be obtained from $A$ by applying to every object and morphism in that diagram the functor $F$). However, in general, the converse is not true: functors don't necessarily reflect diagrams.

What is true is that equivalences of categories both preserve and reflect diagrams, i.e., one can without problems translate back and forth between the categories $C$ and $D$.

I wonder: Given a pair of adjoint functors (say $F\dashv U$, where $F\colon C\to D$ and $U\colon D\to C$) between $C$ and $D$, how much can we translate back and forth between $C$ and $D$?

Through a previous thread I came to the following conclusion: a diagram \begin{array}{cc} \,\,\,\,\,\,\,A \\ \\ i \downarrow & \,\,\,\,\,\,\,\searrow \,{f} \\ \\ A' & \xrightarrow{r} & UB \end{array} in $C$ commutes if and only if the diagram \begin{array}{cc} \,\,\,\,\,\,\,FA \\ \\ Fi \downarrow & \,\,\,\,\,\,\,\searrow \,{\alpha(f)} \\ \\ FA' & \xrightarrow{\alpha(r)} & B \end{array} commutes in $D$, where $\alpha$ denotes the natural bijection $\hom(-, UB)\cong \hom(F-,B)$.

I admit this is not exactly "reflection of diagrams" in the above sense, since only on one arrow we apply $F$ while on the other two we apply $\alpha$.

Question: Can this observation be generalized? Which kind of diagrams are preserved and reflected by adjunctions and in which sense?


1 Answer 1


Instead of talking about reflecting diagrams, let's talk about reflecting equality of arrows. After all, to say a diagram commutes is to say that any two paths through it (i.e., two arrows from one object in the diagram to another) are equal. To say that a functor $F:C\to D$ reflects equality of arrows is to say that each map $F\colon \mathrm{Hom}_C(A,B)\to \mathrm{Hom}_D(F(A),F(B))$ is injective. That is, I think you're asking when adjoint functors are faithful.

One answer is that the left adjoint $F$ is faithful when the unit $\eta\colon \text{Id}_C\to UF$ is a monomorphism, and the right adjoint $G$ is faithful when the counit $\varepsilon\colon FU\to \text{Id}_D$ is an epimorphism. And these conditions are frequently satisfied!

Given arrows $f,g\colon A\to B$ in $C$, such that $F(f) = F(g)$, we also have $GF(f) = GF(g)$, and we get a naturality square: $$\require{AMScd} \begin{CD} A @>{f,g}>> B\\ @V{\eta_A}VV @VV{\eta_B}V \\ GF(A) @>{GF(f) = GF(g)}>> GF(B) \end{CD}$$ Now $\eta_B\circ f = GF(f)\circ \eta_A = GF(g) \circ \eta_B = \eta_B\circ g$, so if $\eta_B$ is monic, we get $f = g$.

The argument for the counits is dual, using the naturality square: $$\begin{CD} FG(A) @>{FG(f)=FG(g)}>> FG(B)\\ @V{\varepsilon_A}VV @VV{\varepsilon_B}V \\ A @>{f,g}>> B \end{CD}$$

In the general case, we still get that if $F(f) = F(g)$, then $f$ and $g$ are coequalized by $\eta_B$, and if $G(f) = G(g)$, then $f$ and $g$ are equalized by $\varepsilon_A$.

  • $\begingroup$ I'm not sure faithfulness is exactly the question here. The example in the problem holds for any adjunction, right? $\endgroup$ Commented Nov 9, 2021 at 16:46
  • 2
    $\begingroup$ @KevinArlin True, I guess I answered the question "which adjoints reflect diagrams?", not "which diagrams are reflected by adjoints?". But the question as stated is fairly open-ended, and I think the observations in my answer might be useful/interesting to the OP. $\endgroup$ Commented Nov 9, 2021 at 21:30
  • $\begingroup$ @AlexKruckman Thanks, but your observation was already clear for me when asking the question. Even if I my question is open-ended, I clearly asked about arbitrary adjunctions and not fully faithful functors. And also about a generalization of an observation from the previous thread I mentioned. Your answer doesn't address any of these main points. $\endgroup$
    – user984603
    Commented Nov 12, 2021 at 15:03

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