# What is the left adjoint for taking rows of a bisimplicial set?

Let $aSS$ / $abSS$ be the category of augmented bi/simplicial sets (one can also consider $SS$/$bSS$ be the usual bi/simplicial sets, the results should related in some reasonable way.)

There is an obvious functor $n$-th row $R_n$ sending an augmented bisimplicial set $X_{\bullet,\bullet}$ to $X_{n,\bullet}$. What is its left adjoint? (exists?)

(In general, how to the compute the left adjoint functor to such a functor $f^*\colon Psh(B)\to Psh(A)$ induced by $f\colon A\to B$. The right adjoint seems easy, if I am not wrong.)

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The left and right adjoints of $f^* : \mathrm{Psh}(B) \to \mathrm{Psh}(A)$ are called Kan extensions. You can find the details in [Categories for the working mathematician]. – Zhen Lin Oct 27 '13 at 2:35

I'll answer the second question, as it subsumes the first. Indeed, if $[n]: 1 \to \Delta$ names the simplex of dimension $n$, then the $n^{th}$-row functor of the question is $f^\ast$ where $f = \Delta \simeq 1 \times \Delta \stackrel{[n] \times 1_\Delta}{\to} \Delta \times \Delta$.

If $A$ and $B$ are small categories and $f: A \to B$ is a functor, then the functor $f^\ast: Psh(B) \to Psh(A)$ obtained by "pulling back" or composing along $f$ has both a left and right adjoint. Since you want the left adjoint, here is a general recipe.

Each presheaf $X: A^{op} \to Set$ can be canonically expressed as a colimit of representables; this is often done in terms of coends:

$$X \cong \int^{a \in Ob(A)} X(a) \cdot A(-, a)$$

where $S \cdot F$ means an $S$-indexed coproduct of copies of $F$, and $A(-, a)$ denotes a representable. The left adjoint to $f^\ast$ takes $X$ to the colimit

$$f_!(X) := \int^{a \in Ob(A)} X(a) \cdot B(-, f(a)).$$

Indeed, for presheaves $Y: B^{op} \to Set$, there is a natural bijection between

• Natural transformations $f_!(X) \to Y$;

• Families of maps $X(a) \cdot B(b, f(a)) \to Y(b)$ that are extranatural in $a$ and natural in $b$;

• Families of maps $X(a) \to \hom(B(b, f(a)), Y(b))$ natural in $a$ and extranatural in $b$;

• Families of maps $X(a) \to Y(f(a))$ natural in $a$ (by applying the Yoneda lemma);

• Natural transformations $X \to f^\ast Y$.

Returning to the first question, if $X$ is a simplicial set, then the left adjoint to the $n^{th}$ row functor takes $X$ to the (augmented) bisimplicial set whose value at $(p, q)$ is

$$\int^{m \in \Delta} X(m) \cdot (\Delta \times \Delta)((p, q), (n, m)) \cong \int^m X(m) \times \Delta(p, n) \times \Delta(q, m) \cong \Delta(p, n) \times X(q).$$

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So the `box product' with $\Delta^n_+\boxtimes ( )$ is the left adjoint, thanks a lot. – Ma Ming Oct 26 '13 at 22:25