I am filling in the details of a proof of this in MacLane, which uses the lim/$\Delta $ adjunction:

Suppose $F\dashv G:X\rightleftharpoons A$, let $J$ be an (index) category, and let $T:J\rightarrow A$ have limit $L$ in $A$ with limit cone ($L,\tau $)

Passing to the functor categories, we get an adjunction $F^{J}\dashv G^{J}:X^{J}\rightleftharpoons A^{J}$ with the functors and units/counits defined in the obvious way:

$F^{J}S=FS, \eta ^{J}(S)=\eta_{S}$ etc.

It then follows trivially that $F^{J}\circ \Delta =\Delta \circ F$ and now, using composites, we get that lim$\circ G^{J}$ and $G\circ$ lim have the same left adjoint and so are isomorphic. Call this isomorphsim $\phi $.

Evaluating at $T$ shows that

lim$GT\cong GL=G$lim$T$

But this is not enough, of course. I need to show $G$ preserves the limit cone and I want to use only the lim/$\Delta $ adjunction, for which I know that $ \tau$ is the $T$ component of the counit. Now, if (lim$GT,\sigma _{i} $) is a limit cone for $GT$, then ($GL,\sigma _{i}\circ \phi)$ is also a limit cone for $GT$ and I want to show that $\sigma _{i}\circ \phi_{T}=G\tau _{i}$, using only the lim/$\Delta $ adjunction.

Final edit: It was a real slog.

1). $F^{J}\circ \Delta \dashv $lim$\circ G^{J}$ and the counit $E$ is $\epsilon ^{J}\circ F^{J}\epsilon ^{\circ }G^{J}$ where $\epsilon ^{\circ }$ is the counit of the $\Delta$/lim adjunction.

2). $\Delta \circ F\dashv G\circ $lim and the counit $\overline E$ is $\epsilon ^{\circ }\circ \Delta \epsilon$lim.

3). Evaluating these at $T\in A^{J}$, and then at $i\in J$ and simplifying, we get $\overline E_{T}(i)=\tau _{i}\circ \epsilon _{L}$ and $E_{T}(i)=\epsilon _{T(i))}\circ F\sigma _{i}$.

4). The isomporphism $\phi $ satisfies $E\circ \Delta F\phi =\overline E$. Evaluating as before, on $T$ then $i$ we find that $\epsilon _{T(i)}\circ F(\sigma _{i}\circ \phi _{T})=\tau _{i}\circ \epsilon _{L}$.

5). To this last result, apply $G$ and then precompose with $\eta _{GL}$, to find, after using a triangle identity, that $G(\epsilon _{T(i)}\circ F(\sigma _{i}\circ \phi_{T})\circ \eta _{GL}=G\tau _{i}$. Set the term in parentheses equal to $h$.Then $\Phi (h)=G\tau _{i}$, where $\Phi $ is the adjunction isomorphism on hom-sets for $F\dashv G$

6). On the other hand, we also have $\epsilon _{T(i)}\circ F(\sigma _{i}\circ \phi _{T})=h$ so that $\Phi (h)$ is also equal to $\sigma _{i}\circ \phi _{T}$, so we are done.

  • 4
    $\begingroup$ But don't forget that right adjoints preserve limits also in categories which are not complete, although the $\lim-\Delta $-adjunction only exists in complete categories. $\endgroup$
    – user158047
    Nov 17 '14 at 5:47
  • $\begingroup$ @Jakob: Right. In fact, although Maclane calls this approach "more sophisticated", it is less general, and harder than the four line direct proof. $\endgroup$ Nov 17 '14 at 15:05
  • 1
    $\begingroup$ This is only a technical problem. If necessary, pass to the universal completion, and in the end you will land in your category again. $\endgroup$ Nov 17 '14 at 15:55
  • $\begingroup$ Yes, but MacLane should mention this. $\endgroup$
    – user158047
    Nov 17 '14 at 16:52
  • $\begingroup$ It still seems like a lot of work to get a result that follows easily from the direct use of the original adjunction. $\endgroup$ Nov 18 '14 at 1:20

When you go through the proof in detail you will see that $GL \to \lim GT \to G T_j$ is induced by the cone $L \to T_j$.

  • $\begingroup$ this is exactly what do not see how to do. Can you give me a hint on how to do this, using the construction of the functor categories as above and the lim/$\Delta $ adjunction? I will post my edit in the main problem, to say what I have so far. Thanks. $\endgroup$ Nov 17 '14 at 15:31
  • 1
    $\begingroup$ You have used two facts: 1) Adjoints are unique. 2) Adjoints of compositions are compositions of adjoints. Write down their proofs. Apply these proofs here. Then you will be done. (Sorry, I'm a bit too lazy to do that right now.) $\endgroup$ Nov 17 '14 at 15:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.