If a closed monoidal category is the monoidal equivalent of a Cartesian closed category, is there an analogous equivalent for locally cartesian closed categories? Is there a standard terminology or reference for such a construction?

5$\begingroup$ What would you want such a thing to do? The first problem that occurs to me is that I don't see why the slice categories of a monoidal category should even have a monoidal structure. $\endgroup$ – Qiaochu Yuan May 6 '15 at 19:46

1$\begingroup$ Well, if we take "slices" to be part of the data – that is to say, if we start with an indexed (or fibred) category – then we could certainly speak of indexed monoidal categories and indexed monoidal closed categories. $\endgroup$ – Zhen Lin May 6 '15 at 20:22

$\begingroup$ @Qiaochu  I am coming from a context of categorical logic, where we use pullbacks for everything (substitution, conjunction, weakening). Other operations are usually assumed to be pullbackstable as a matter of course. Now I am running into monoidal categories and I would like to know to what extent those manipulations can translate into a nonCartesian context. $\endgroup$ – pnips May 6 '15 at 20:44

5$\begingroup$ @pnips Then you should look into linear logic. $\endgroup$ – Zhen Lin May 6 '15 at 22:26
In a certain sense a monoidal version of a slice category is a category of comodules over a cocommutative comonoid object. If $C$ is a cocommutative comonoid object in a monoidal category, then the category of comodules over it $\mathbf{Comod}(C)$ has a monoidal structure with the monoidal product defined in a standard way as a tensor product $\otimes_C$ over $C$. In the Cartesian case a comonoid is just an object, and $\mathbf{Comod}(C)$ coincides with the slice category.
From this point of view a "monoidal equivalent of a locally Cartesian closed category" is a monoidal category for which all $\mathbf{Comod}(C)$ are monoidal closed.
Locally Cartesian closed categories are examples. Another example is the opposite of the category of abelian groups $\mathbf{Ab}^\mathrm{op}$, since the category of modules $\mathbf{Mod}(R)$ over any commutative ring $R$ is a closed monoidal category (it is probably locally coclosed).

1$\begingroup$ If $C$ is a comonoid in $Ab^{op}$ (= monoid in $Ab$), then the category of comodules over $C$ is the opposite of the category of modules over $C$. The category of modules is closed, but its opposite is not, so $Ab^{op}$ is not an example of a locally monoidal closed category. $\endgroup$ – Valery Isaev Jan 13 '17 at 16:19

1$\begingroup$ So, are there any interesting examples of locally monoidal closed categories besides cartesian? $\endgroup$ – Valery Isaev Jan 13 '17 at 16:22
I think there is something intriguing and slightly mysterious going on here.
First, my proposed definition would be slightly different from Dimitri Chikladze's. I agree that the natural generalization of the slice categories $\mathcal V /C$ for $C \in \mathcal V$ should probably be the comodule categories $\mathbf{Comod}(C)$ for $C$ a commutative comonoid in $\mathcal V$. But to me, the fundamental way to characerize local cartesian closedness of $\mathcal V$ is that for $f: C \to D$ in $\mathcal V$, the reindexing functor $f_! : \mathcal V/C \to \mathcal V/D$ (defined by postcomposition) not only has a right adjoint $f^\ast$ (defined by pullback), but $f^\ast$ itself has a right adjoint $f_\ast$ (the local hom functor). So I would say:
Definition: A symmetric monoidal category $\mathcal V$ is locally closed if for every commutative comonoid homomorphism $f: C \to D$ in $\mathcal V$, the reindexing functor $f_!$ fits into an adjoint string $f_! \dashv f^\ast \dashv f_\ast$ satisfying BeckChevalley conditions.
The strange thing here is when we look at examples. If $\mathcal V$ has certain nice limits and colimits (e.g. $\mathcal V = \mathsf{Ab}^{\mathrm{op}}$), then we always have an adjoint string $f^! \dashv f_! \dashv f^\ast$ (where $f^\ast$ and $f^!$ are respectively induction and coinduction of modules if $\mathcal V = \mathsf{Ab}^\mathrm{op}$). But the extra adjoint $f^!$ is on the wrong side! In order for $f^\ast$ to have a right adjoint, it must be flat when viewed as a monoid homomorphism in $\mathcal V^\mathrm{op}$.
I can't think of a noncocartesian example of a $\mathcal V^\mathrm{op}$ for which all commutative monoid homomorphisms are flat  even $\mathsf{Vect}_k$ for $k$ a field doesn't seem to work. So it looks like in noncartesian cases, we're already doing what we do with cartesian categories like $\mathsf{Cat}$  exponentiability of a morphism is an important property that one is often interested in, but it's just not reasonable to expect every morphism to be exponentiable.
But perhaps there are natural subcategories of all commutative comonoids which have better exponentiability properties in some cases.

$\begingroup$ @HarryGindi That would be really cool if true. I would be careful, though, given that the cartesian product is not exponentiable on Cat or qCat. In that case, you can define a local hom at the level of simplicial sets, but it doesn't play well with Joyal fibrant replacement. So there are issues to consider... $\endgroup$ – Tim Campion Jun 10 '18 at 21:44