ScalingStacks

Proof. The claim is that for any $k$-correspondence $N\to C_{k}$, the functor

$$-\times_{C_{k}}N\colon(\gaunt_{n}/C_{k})\to(\gaunt_{n}/C_{k})$$

admits a right adjoint $\uHom_{C_{k}}(N,-)$.

Since $\gaunt_{n}/C_{k}$ is locally finitely presentable, it is cocomplete and admits a strong generator [1, Th. 1.20] and is co-wellpowered [1, Th. 1.58]. Thus by the special adjoint functor theorem [30, Sect. V.8] the functor

$$-\times_{C_{k}}N\colon(\gaunt_{n}/C_{k})\to(\gaunt_{n}/C_{k})$$

admits a right adjoint precisely if it commutes with colimits.

When $k=0$, this follows from the fact that $\gaunt_{n}$ itself is cartesian closed.

For $k>0$, suppose $(\gaunt_{n-1}/C_{k-1})$ is cartesian closed. To prove that the category $(\gaunt_{n}/C_{k})$ is cartesian closed, we require a description of colimits in terms of the equivalent category $\corr_{n}^{k}$. For any small category $\Lambda$ and any diagram $X\colon\Lambda\to\corr_{n}^{k}$ with

$$X_{\lambda}=(X_{\lambda,0},X_{\lambda,1},F_{\lambda}),$$

it is easy to see that the colimit is given by the triple $(X_{0},X_{1},F)$ where

$$X_{0}=\colim_{\lambda\in\Lambda}X_{\lambda,0}\text{\quad and\quad}X_{1}=\colim_{\lambda\in\Lambda}X_{\lambda,1},$$

and $F\colon X_{0}^{\mathrm{op}}\times X_{1}\to\corr_{n-1}^{k-1}$ is the enriched left Kan extension of

$$\colim_{\lambda\in\Lambda}F_{\lambda}\colon\colim_{\lambda\in\Lambda}(X_{\lambda,0}^{\mathrm{op}}\times X_{\lambda,1})\to\corr_{n-1}^{k-1}$$

along the diagonal

$$\colim_{\lambda\in\Lambda}(X_{\lambda,0}^{\mathrm{op}}\times X_{\lambda,1})\to X_{0}^{\mathrm{op}}\times X_{1}.$$

Now for any object $Y=(Y_{0},Y_{1},G)\in\corr_{n}^{k}$, we wish to compare $(\colim_{\lambda\in\Lambda}X_{\lambda})\times Y$ and $\colim_{\lambda\in\Lambda}(X_{\lambda}\times Y)$. In light of our descriptions of products in $\corr_{n}^{k}$, we see that the former is $(X_{0}\times Y_{0},X_{1}\times Y_{1},F\otimes G)$, and the latter is the colimit of the diagram $Z\colon\Lambda\to\corr_{n}^{k}$ that carries $\lambda$ to

$$(X_{\lambda,0}\times Y_{1},X_{\lambda,1}\times Y_{1},F_{\lambda}\otimes G).$$

Note that, since $\gaunt_{n}$ is cartesian closed, one has

$$\colim_{\lambda\in\Lambda}(X_{\lambda,0}\times Y_{0})\cong X_{0}\times Y_{0}\text{\quad and\quad}\colim_{\lambda\in\Lambda}(X_{\lambda,1}\times Y_{1})\cong X_{1}\times Y_{1};$$

hence our description of colimits in $\corr_{n}^{k}$ exhibits the colimit of $Z$ as $(X_{0}\times Y_{0},X_{1}\times Y_{1},(F\otimes G)^{\prime})$, where $(F\otimes G)^{\prime}$ is the enriched left Kan extension of

$$\colim_{\lambda\in\Lambda}(F_{\lambda}\otimes G)\colon\colim_{\lambda\in\Lambda}((X_{\lambda,0}\times Y_{0})^{\mathrm{op}}\times(X_{\lambda,1}\times Y_{1}))\to\corr_{n-1}^{k-1}$$

along the diagonal

$$\colim_{\lambda\in\Lambda}((X_{\lambda,0}\times Y_{0})^{\mathrm{op}}\times(X_{\lambda,1}\times Y_{1}))\to(X_{0}\times Y_{0})^{\mathrm{op}}\times(X_{1}\times Y_{1}).$$

Our induction hypothesis is that $\corr_{n-1}^{k-1}$ is cartesian closed; so this enriched left Kan extension can be identified with the composition of the enriched left Kan extension of

$$\colim_{\lambda\in\Lambda}(F_{\lambda},G)\colon\colim_{\lambda\in\Lambda}(X_{\lambda,0}^{\mathrm{op}}\times X_{\lambda,1})\times(Y_{0}^{\mathrm{op}}\times Y_{1})\to\corr_{n-1}^{k-1}\times\corr_{n-1}^{k-1}$$

along the diagonal

$$\colim_{\lambda\in\Lambda}(F_{\lambda},G)\colon\colim_{\lambda\in\Lambda}(X_{\lambda,0}^{\mathrm{op}}\times X_{\lambda,1})\times(Y_{0}^{\mathrm{op}}\times Y_{1})\to(X_{0}^{\mathrm{op}}\times X_{1})\times(Y_{0}^{\mathrm{op}}\times Y_{1}).$$

But now this left Kan extension is simply the product of $G$ with the left Kan extension that defines $F$. In other words, we have an isomorphism $(F\otimes G)^{\prime}\cong F\otimes G$, whence the proof is complete. ∎