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I'm trying to understand the proof "each functor $D \rightarrow \mathbf{Set}$ is a colimit of representable functors" in MacLane (CFWM), but like with all things Yoneda I'm having problems.

1.) After he takes $L$ to be vertex of some other cone over $M$ (say $\psi$), he says that we have $z$ and $z'$ s.t. $y^{-1} z = \psi_{(d,x)}$ and $y^{-1} z' = \psi_{(d',x')}$. But I lose him in the next part, where he claims that $z' = f z$ (I suppose this means $z' = L(f)(z)$). Why is this so?

2.) He then constructs $\theta_d$ that maps $x \in Kd$ to $z$ such that $y^{-1}z$ is in the cone of $L$. If you were to take $\eta \in D(d', u)$, then the following needs to hold: $$ \theta_u \circ (y^{-1}x')_u (\eta) = (y^{-1}z')_u(\eta) $$ Why does this hold?

Thanks.

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up vote 3 down vote accepted

Note that there is a typo in that version of CWM: The diagram should be $$M:\mathcal{J}^{\mathrm{op}} \to \textbf{Sets}^{\mathcal{D}}$$ instead of $$M:\mathcal{J}^{\mathcal{D}}\to \textbf{Sets}^{\mathcal{D}}$$ If you really want to learn the proof of this fact, I would recommend realizing it as a consequence of the fact that if $\mathcal{C}$ is a small category, $\mathcal{D}$ a cocomplete category and $F:\mathcal{C}\to\mathcal{D}$ a functor, then the functor \begin{eqnarray*} R:\mathcal{D}&\to&[\mathcal{C}^{\text{op}},\mathbf{Sets}]\\ D&\mapsto& R(D) \end{eqnarray*} where \begin{eqnarray*} R(D):\mathcal{C}^{\text{op}}&\to&\mathbf{Sets}\\ C&\mapsto& \text{Hom}_{\mathcal{D}}(F(C),D) \end{eqnarray*} has a left adjoint. Because it is a much more useful result with many other applications. And the proof is really quite similar in spirit to the one in CWM. So, two birds with one stone :)

This is Theorem 2 of Section I of MacLane and Moerdijk's "Sheaves in Geometry and Logic", where the result you are asking for appears as Corollary 3.

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Actually the typo is a bit deeper because the functor M described in the text is contravariant. What I would have preferred is if $\mathcal J$ was defined as $\mathcal J \overset{def}= (\mathbf 1 \downarrow K)^{\mathrm{op}}$ since cones are defined in terms of covariant functors, so that this new definition of $\mathcal J$ becomes the desired diagram ; but a quick dirty way to fix everything is to just write $\mathcal J^{\mathrm{op}}$, consider the diagram as being $\mathcal J^{\mathrm{op}}$ and write $M : \mathcal J^{\mathrm{op}} \to \mathbf{Sets}^D$. – Patrick Da Silva Feb 25 at 16:00

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