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This question is an exact duplicate of:

Theorem : Representable functors preserve limits.

I'm struggling to see why this is true. It's not obvious to me where I should be actually using the fact that functor in question is representable.

Any help would be much appreciated.

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marked as duplicate by Zhen Lin, Qiaochu Yuan, Henry T. Horton, Amzoti, O.L. May 15 '13 at 0:22

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Do you know that $hom(a,-)$ preserve limits? –  Oskar May 14 '13 at 22:42
The hom functor preserves limits then since it's isomorphic to this functor the result follows –  user40276 May 14 '13 at 22:43
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1 Answer

Firstly we have to prove that $hom(a,-)$ preserves limits.

Let $A$ be a category, $a\in A$ be its object and $hom_A(a,-)\colon A\to\mathbf{Set}$ be a hom-functor, corresponding to this object, let $B$ is also a category, $T\colon B\to A$ be a functor. Let's assume that $a_0$ is a limit of $T$, thus, we can consider a limiting cone $\varphi\colon\Delta_{a_0}\to T$. We want to prove that $hom_A(a,a_0)$ is a limit of the functor $hom_A(a,T(-))$ and that the induced cone $\varphi_*\colon\Delta_{hom_A(a,a_0)}\to hom_A(a,T(-))$ is also limiting. Let $X$ be an arbitrary set and $\psi\colon\Delta_X\to hom_A(a,T(-))$ be an arbitrary natural transformation. Now we want to find a mapping $f\colon X\to hom_A(a,a_0)$, such that $\varphi_*\circ\Delta_f=\psi$. For any element $x\in X$, let's construct a natural transformation $\psi_x\colon\Delta_a\to T$, such that $\psi_x(b)=(\psi(b))(x)$(check that this is a natural transformation). This construction will give you an arrow $f_x\colon a\to a_0$, such that $\varphi\circ\Delta_{f_x}=\psi_x$. Define the mapping $f$ by putting $f(x)=f_x$. Check that such mapping is unique which satisfies the property $\varphi_*\circ\Delta_f=\psi$.

Secondly, if $A$ and $B$ are categories, $T,S\colon A\to B$ are naturally isomorphic functors, then $T$ is continious(=preserves all limits) iff $S$ is continious(it is an easy exercise).

Now, if you have a representable functor, then it is isomorphic to some hom-functor, and therefore it is also continious - QED.

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