I'm attempting to prove that every presheaf is a canonical colimit of representable presheaves by constructing a limiting cocone directly (I'm aware that there are more elegant proofs, but this is more of an exercise).

Most of the proof is fine, but I'm having trouble understanding why a certain map is actually natural.

To fix notation, take a small category $\mathbf{C}$ and let $\mathbf{y} : \mathbf{C} \rightarrow \widehat{\mathbf{C}}$ denote the Yoneda embedding and $\int P$ the category of elements of $P$. We intend to show that $P \cong \text{colim} \left( \int P \xrightarrow{\pi_P} \mathbf{C} \xrightarrow{\mathbf{y}} \widehat{\mathbf{C}} \right)$.

Omitting most of the details, let $\left( h_X \xrightarrow{\alpha^{X,x}} P \right)_{(X, x) \in \int P}$ be the limit cocone. ($\alpha^{X,x}$ is just defined to be the natural transformation corresponding to $x$ under the Yoneda correspondence), and let $\left( h_X \xrightarrow{\beta^{X,x}} Q \right)_{(X,x) \in \int P}$ be another cone.

One reference defines the universal map $\gamma : P \rightarrow Q$ by letting $\gamma_X(x)$ be the precisely the element of $Q(X)$ that $\beta^{X,x}$ corresponds to under the Yoneda correspondence. Explicitly, $\gamma_X(x) = \beta^{X,x}_X(1_X)$.

This seems perfectly natural (as in "no arbitrary choices") but I fail to see rigorously why it is a natural transformation. We get the following diagrams for some $f : X \rightarrow Y$.

enter image description here

And I fail to see why the two values in the top right should be equal. I've tried writing down every commutative diagram I can think of at my disposal involving those actors but I can't seem to get the right subscripts in the right places. It's not seeming straightforward why this transformation should be natural.

Of course, like all things Yoneda related, it probably falls out of something simple and fundamental that I'm simply missing. What is the proper reasoning to show that this transformation is natural?

Finally, I'm not sure why the map $\gamma$ is unique. It's easy to show that $\gamma \circ \alpha^{X,x} = \beta^{X,x}$, but to show that $\gamma$ is unique would mean to show that $\beta^{X,x} = \gamma \circ \alpha^{X,x} = \delta \circ \alpha^{X,x}$ implies that $\gamma = \delta$, which amounts to showing that $\alpha^{X,x}$ is an epimorphism, which does not seem to be an automatic conclusion. Why should this map be unique?

  • $\begingroup$ You need to use the fact that each of your cocones is... a cocone, and that $\beta^{X,x}$ is a natural transformation. For unicity, you have $\gamma\circ \alpha^{X,x}=\delta\circ \alpha^{X,x}$ for all $X,x$ ! $\endgroup$ – Max Jul 24 '18 at 8:57

So what you want is to prove that $$\beta_X^{X,P(f)(y)}(1_X)=Q(f)(\beta_Y^{Y,y}(1_Y)).\tag{1}\label{1}$$ For this, you can use the fact that the maps $(\beta^{X,x})_{(X,x)\in \int P}$ form a cocone, which tells you that, since $f$ defines a map $(X,P(f)(y))\to (Y,y)$ in $\int P$, you have $\beta^{X,P(f)(y)}=\beta^{Y,y}\circ \mathbf{y}(f)$. In particular, the left-hand side in \eqref{1} is equal to $$\beta_X^{Y,y}(\mathbf{y}(f)_X(1_Y))=\beta_X^{Y,y}(f).$$

Moreover, you can use the fact that $\beta^{Y,y}:h_X=\mathbf{C}(\_,X)\Rightarrow Q$ is natural : this tells you in particular that $Q(f)\circ \beta^{Y,y}_Y=\beta^{Y,y}_X\circ h_Y(f)$, and thus the right-hand side of \eqref{1} is equal to $$\beta^{Y,y}_X(h_Y(f)(1_Y))=\beta_X^{Y,y}(f).$$ This confirms that the two sides are equal.

As for the uniqueness, assume that $\delta$ is a natural transformation $P\Rightarrow Q$ such that $\delta\circ\alpha^{X,x}=\beta^{X,x}$ for all $(X,x)\in \int P$. Note that, by definition of the cocone $(\alpha^{X,x})_{(X,x)\in \int P}$ and thanks to the Yoneda lemma, we have $\alpha^{X,x}_X(1_X)=x$ for all $X$ and all $x\in P(X)$. In particular, $$\delta_X(x)=\delta_X(\alpha^{X,x}_X(1_X))=\beta_X^{X,x}(1_X)=\gamma_X(x),$$ hence $\delta=\gamma$.

  • $\begingroup$ Ah, I had tried combining the naturality square and the cocone triangle for the $\beta$'s but I hadn't thought of meeting in the middle like that! As for the uniqueness, I think I was just looking at it from the wrong angle - your argument is quick and simple. Thanks! $\endgroup$ – cemulate Jul 24 '18 at 16:51

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