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Suppose I have an adjunction $\mathcal{C}\overset{R}{\underset{I}\leftrightarrows}\mathcal{D}$, where $R\dashv I$, and $I$ is full and faithful. Now let $F:\mathcal{A}\rightarrow\mathcal{C}$ be any diagram, I then want to prove that if $IF$ has a limit in $\mathcal{D}$, then $F$ has a limit in $\mathcal{C}$.

I suppose I have to start by choosing an arbitrary limit of $IF$ in $\mathcal{D}$ and from that construct a limit of $F$ in $\mathcal{C}$. However, I am not really sure where to go from there.

Any help would be appriciated.

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$I$ preserves the limit. So there is an obvious candidate... – Zhen Lin Mar 23 '14 at 9:46
But doesn't this only really help me the other way? Doesn't this mean that if $F$ has a limit, then so does $IF$, but not that if $IF$ does, that $F$ does? – matti0006 Mar 23 '14 at 9:53
That way around is not true in general. I am going to guess that $I$ is supposed to be fully faithful – then it is true. – Zhen Lin Mar 23 '14 at 10:40
Yes, you are right, $I$ is full and faithful, however, I still don't see how the fact that $I$ preserves limits helps me. Could you throw me another hint perhaps? – matti0006 Mar 23 '14 at 10:44
I also figured this out: If $IF(A)$ is a limit in $\mathcal{D}$ then by adjunction, for any object $D$ in $\mathcal{D}$ there is a bijection between $\mathcal{D}(D,IF(A))$ and $\mathcal{C}(R(D),F(A))$. Since $IF(A)$ is a limit there is a unique arrow to it from every $D$ in $\mathcal{D}$ so there also is a unique arrow from every $R(D)$ to $F(A)$. This seems to almost complete the proof but I am left with whether $R$ is neccesarily surjective. – matti0006 Mar 23 '14 at 10:48

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