# Right Ajoints Preserve Limits--alternate proof

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.

• 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.
– user158047
Nov 17 '14 at 5:47
• @Jakob: Right. In fact, although Maclane calls this approach "more sophisticated", it is less general, and harder than the four line direct proof. Nov 17 '14 at 15:05
• This is only a technical problem. If necessary, pass to the universal completion, and in the end you will land in your category again. Nov 17 '14 at 15:55
• Yes, but MacLane should mention this.
– user158047
Nov 17 '14 at 16:52
• It still seems like a lot of work to get a result that follows easily from the direct use of the original adjunction. 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 $L \to T_j$.
• 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. Nov 17 '14 at 15:31