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Let $\mathbb{C}$ be a small category and $X\in \mathbb{C}$ an object. The category $\widehat{\mathbb{C}}$ of presheaves on $\mathbb{C}$ is cartesian closed, i.e. each product \begin{equation} -\times A:\widehat{\mathbb{C}}\rightarrow \widehat{\mathbb{C}} \end{equation} has a right adjoint $R_A:\widehat{\mathbb{C}}\rightarrow \widehat{\mathbb{C}}$ given by $R_A(B)(C)=Hom_{\widehat{\mathbb{C}}}(A\times Y(C), B)$ where $Y$ is the Yoneda embedding.

In the answer of Zhen Lin to this other question, an isomorphism \begin{equation} \widehat{\mathbb{C}}\downarrow Y(X) \cong \widehat{\mathbb{C}\downarrow X} \end{equation} is constructed. This gives an easy description of the right adjoint $R_A^X$ to each product \begin{equation} -\times_X A:\widehat{\mathbb{C}}\downarrow Y(X)\rightarrow \widehat{\mathbb{C}}\downarrow Y(X) \end{equation} of the over category by $R_A^X(B)(C)=Hom_{\widehat{\mathbb{C}}\downarrow Y(X)}(A\times_X Y(C),B)$. This notion is a little messy because $Y$ isn't the same Yoneda embedding as before, $A$ and $B$ aren't the same as before because I use the isomorphism but this does not matter for my question.

I want to get a description of this ''exponential'' of the over category without using the above isomorphism, i.e. as a presheaf on $\mathbb{C}$ over $Y(X)$. As far as I understand, Proposition 1 of this nLab entry defines something like \begin{eqnarray} R^X_{a:A\to X}(B\to X)(C)&=&Hom_{\widehat{\mathbb{C}}\downarrow Y(A)}(a^*(C),a^*(B\to X))\\ &=& Hom_{\widehat{\mathbb{C}}\downarrow Y(A)}(a^*(C),A\times_X B\to A) \end{eqnarray} with $a^*:\widehat{\mathbb{C}}\downarrow Y(X)\rightarrow \widehat{\mathbb{C}}\downarrow Y(A)$ but I did something wrong here: How can I apply $a^*$ to $C\in\mathbb{C}$? Also I don't see, how this is given a morphism to $Y(X)$.

My question is: How I can I understand the right adjoint to the product $-\times_X A$ in the over category $\widehat{\mathbb{C}}\downarrow Y(X)$ as a functor from $\mathbb{C}^{op}$ to $Set$ over $Y(X)$?

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I don't understand the point of your question. If it is easier to describe the construction in $\widehat{\mathbb{C} \downarrow X}$ than in $\hat{\mathbb{C}} \downarrow Y(X)$, then why not use the construction in $\widehat{\mathbb{C} \downarrow X}$? No matter. Let's just translate what's going on in the first description to the second description.

Given objects $A \to Y(X)$ and $B \to Y(X)$ in $\hat{\mathbb{C}} \downarrow Y(X)$, we construct the fibred exponential as follows: $$B^A (C) = \coprod_{c \in \mathbb{C}(C, X)} (\hat{\mathbb{C}} \downarrow Y(X)) (\tilde{Y}( c) \times_{Y (X)} A, B)$$ Here, $\tilde{Y}$ is the modified Yoneda embedding $(\mathbb{C} \downarrow X) \to (\hat{\mathbb{C}} \downarrow Y(X))$ given by $$\tilde{Y} (c) (D) = \coprod_{d \in \mathbb{C}(D, X)} (\mathbb{C} \downarrow X)(c, d)$$ The structural morphisms $B^A \to Y(X)$ and $\tilde{Y}(c) \to Y(X)$ are the obvious ones arising from their fibrewise construction.

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  • $\begingroup$ Thank for answering. The point is understanding what's going on. $\endgroup$ Sep 1, 2012 at 7:17

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