The slice category or over category $\mathbf{C}/c$ of a category $\mathbf{C}$ over an object $c \in \mathbf{C}$ has
objects that are all arrows $f \in \mathbf{C}$ such that $cod(f) = c$, and
morphisms $g: X \to X' \in \mathbf{C}$ from $f:X \to c$ to $f': X' \to c$ such that $f' \circ g = f$.
The slice category is a special case of a comma category.
There is a forgetful functor $U_c: \mathbf{C}/c \to \mathbf{C}$ which maps an object $f:X \to c$ to its domain $X$ and a morphism $g: X \to X' \in \mathbf{C}/c$ (from $f:X \to c$ to $f': X' \to c$ such that $f' \circ g = f$) to the morphism $g: X \to X'$.
The dual notion is an under category.
If $\mathbf{C} = \mathbf{P}$ is a poset and $p \in \mathbf{P}$, then the slice category $\mathbf{P}/p$ is the down set $\downarrow (p)$ of elements $q \in \mathbf{P}$ with $q \leq p$.
If $1$ is a terminal object in $\mathbf{C}$, then $\mathbf{C}/1$ is isomorphic to $\mathbf{C}$.
For $X$ a topological space then the category of covering spaces over $X$ is a full subcategory of the slice category $Top_{/X}$ of the category of topological spaces.
The assignment of overcategories $C/c$ to objects $c \in C$ extends to a functor
Under the Grothendieck construction this functor corresponds to the codomain fibration
from the arrow category of $C$. (Note that unless $C$ has pullbacks, this functor is not actually a fibration, though it is always an opfibration.)
Let
be a pair of adjoint functors, where the category $C$ has all pullbacks.
Then for every object $X \in C$ there is induced a pair of adjoint functors between the slice categories
where
Let $C$ be a category, $c$ an object of $C$ and let $C/c$ be the over category of $C$ over $c$. Write $PSh(C/c) = [(C/c)^{op}, Set]$ for the category of presheaves on $C/c$ and write $PSh(C)/Y(c)$ for the over category of presheaves on $C$ over the presheaf $Y(c)$, where $Y : C \to PSh(c)$ is the Yoneda embedding.
There is an equivalence of categories
The functor $e$ takes $F \in PSh(C/c)$ to the presheaf $F' : d \mapsto \sqcup_{f \in C(d,c)} F(f)$ which is equipped with the natural transformation $\eta : F' \to Y(c)$ with component map $\eta_d \sqcup_{f \in C(d,c)} F(f) \to C(d,c)$.
A weak inverse of $e$ is given by the functor
which sends $\eta : F' \to Y(C))$ to $F \in PSh(C/c)$ given by
where $F'(d)|_c$ is the pullback
Suppose the presheaf $F \in PSh(C/c)$ does not actually depend on the morphsims to $C$, i.e. suppose that it factors through the forgetful functor from the over category to $C$:
Then $F'(d) = \sqcup_{f \in C(d,c)} F(f) = \sqcup_{f \in C(d,c)} F(d) \simeq C(d,c) \times F(d)$ and hence $F ' = Y(c) \times F$ with respect to the closed monoidal structure on presheaves.
See also functors and comma categories.
For the analogous statement in (∞,1)-category theory see (∞,1)-category of (∞,1)-presheaves – Interaction with overcategories
at (∞,1)-category of (∞,1)-presheaves.
A limit in an under category is computed as a limit in the underlying category.
Precisely: let $C$ be a category, $t \in C$ an object, and $t/C$ the corresponding under category, and $p : t/C \to C$ the obvious projection.
Let $F : D \to t/C$ be any functor. Then, if it exists, the limit over $p \circ F$ in $C$ is the image under $p$ of the limit over $F$:
and $\lim F$ is uniquely characterized by $\lim (p F)$.
Over a morphism $\gamma : d \to d'$ in $D$ the limiting cone over $p F$ (which exists by assumption) looks like
By the universal property of the limit this has a unique lift to a cone in the under category $t/C$ over $F$:
It therefore remains to show that this is indeed a limiting cone over $F$. Again, this is immediate from the universal property of the limit in $C$. For let $t \to Q$ be another cone over $F$ in $t/C$, then $Q$ is another cone over $p F$ in $C$ and we get in $C$ a universal morphism $Q \to \lim p F$
A glance at the diagram above shows that the composite $t \to Q \to \lim p F$ constitutes a morphism of cones in $C$ into the limiting cone over $p F$. Hence it must equal our morphism $t \to \lim p F$, by the universal property of $\lim p F$, and hence the above diagram does commute as indicated.
This shows that the morphism $Q \to \lim p F$ which was the unique one giving a cone morphism on $C$ does lift to a cone morphism in $t/C$, which is then necessarily unique, too. This demonstrates the required universal property of $t \to \lim p F$ and thus identifies it with $\lim F$.
One often says “$p$ reflects limits” to express the conclusion of this proposition. A conceptual way to consider this result is by appeal to a more general one: if $U: A \to C$ is monadic (i.e., has a left adjoint $F$ such that the canonical comparison functor $A \to (U F)-Alg$ is an equivalence), then $U$ both reflects and preserves limits. In the present case, the projection $p: A = t/C \to C$ is monadic, is essentially the category of algebras for the monad $T(-) = t + (-)$, at least if $C$ admits binary coproducts. (Added later: the proof is even simpler: if $U: A \to C$ is the underlying functor for the category of algebras of an endofunctor on $C$ (as opposed to algebras of a monad), then $U$ reflects and preserves limits; then apply this to the endofunctor $T$ above.)
For $\mathcal{C}$ a category, $X \;\colon\; \mathcal{D} \longrightarrow \mathcal{C}$ a diagram, $\mathcal{C}_{/X}$ the comma category (the over-category if $\mathcal{D}$ is the point) and $F \;\colon\; K \to \mathcal{C}_{/X}$ a diagram in the comma category, then the limit $\underset{\leftarrow}{\lim} F$ in $\mathcal{C}_{/X}$ coincides with the limit $\underset{\leftarrow}{\lim} F/X$ in $\mathcal{C}$.
For a proof see at (∞,1)-limit here.
As a special case of the above discussion of limits and colimits in a slice $\mathcal{C}_{/X}$ we obtain the following statement, which of course is also immediately checked explicitly.
If $\mathcal{C}$ has an initial object $\emptyset$, then $\mathcal{C}_{/X}$ has an initial object, given by $\langle \emptyset \to X\rangle$.
The terminal object of $\mathcal{C}_{/X}$ is $\mathrm{id}_X$.
Last revised on August 12, 2019 at 12:38:04. See the history of this page for a list of all contributions to it.