The Stacks project

Proof. During the proof of this lemma we denote $\text{cosk}_ mU$ the simplicial object with $(\text{cosk}_ mU)_ n$ equal to $\mathop{\mathrm{lim}}\nolimits _{(\Delta /[n])_{\leq m}^{opp}} U(n)$. We will conclude at the end of the proof that it does satisfy the required mapping property.

Suppose that $V$ is a simplicial object. A morphism $\gamma : V \to \text{cosk}_ mU$ is given by a sequence of morphisms $\gamma _ n : V_ n \to (\text{cosk}_ mU)_ n$. By definition of a limit, this is given by a collection of morphisms $\gamma (\alpha ) : V_ n \to U_ k$ where $\alpha $ ranges over all $\alpha : [k] \to [n]$ with $k \leq m$. These morphisms then also satisfy the rules that

are commutative, given any $0 \leq k, k' \leq m$, $0 \leq n, n'$ and any $\psi : [k] \to [k']$, $\varphi : [n] \to [n']$, $\alpha : [k] \to [n]$ and $\alpha ' : [k'] \to [n']$ in $\Delta $ such that $\varphi \circ \alpha = \alpha ' \circ \psi $. Taking $n = k = k'$, $\varphi = \alpha '$, and $\alpha = \psi = \text{id}_{[k]}$ we deduce that $\gamma (\alpha ') = \gamma (\text{id}_{[k]}) \circ V(\alpha ')$. In other words, the morphisms $\gamma (\text{id}_{[k]})$, $k \leq m$ determine the morphism $\gamma $. And it is easy to see that these morphisms form a morphism $\text{sk}_ m V \to U$.

Conversely, given a morphism $\gamma : \text{sk}_ m V \to U$, we obtain a family of morphisms $\gamma (\alpha )$ where $\alpha $ ranges over all $\alpha : [k] \to [n]$ with $k \leq m$ by setting $\gamma (\alpha ) = \gamma (\text{id}_{[k]}) \circ V(\alpha )$. These morphisms satisfy all the displayed commutativity restraints pictured above, and hence give rise to a morphism $V \to \text{cosk}_ m U$. $\square$

Proof. This is true because the category $(\Delta /[n])_{\leq m}$ has an final object in this case, namely the identity map $[n] \to [n]$. $\square$

Proof. Combine Lemmas 14.19.2 and 14.19.3. $\square$

Proof. The limit, if it exists, represents the functor that associates to an object $T$ the set

In fact we will show this functor is isomorphic to the one displayed in the lemma. The map in one direction is given by the rule

This satisfies the conditions of the lemma because

by the relations we recalled above the lemma. To construct a map in the other direction we have to associate to a system $(f_0, \ldots , f_{n + 1})$ as in the displayed formula of the lemma a system of maps $f_\alpha $. Let $\alpha : [k] \to [n + 1]$ be given. Since $k \leq n$ the map $\alpha $ is not surjective. Hence we can write $\alpha = \delta ^{n + 1}_ i \circ \psi $ for some $0 \leq i \leq n + 1$ and some $\psi : [k] \to [n]$. We have no choice but to define

Of course we have to check that this is independent of the choice of the pair $(i, \psi )$. First, observe that given $i$ there is a unique $\psi $ which works. Second, suppose that $(j, \phi )$ is another pair. Then $i \not= j$ and we may assume $i < j$. Since both $i, j$ are not in the image of $\alpha $ we may actually write $\alpha = \delta ^{n + 1}_{i, j} \circ \xi $ and then we see that $\psi = \delta ^ n_{j - 1} \circ \xi $ and $\phi = \delta ^ n_ i \circ \xi $. Thus

as desired. We still have to verify that the maps $f_\alpha $ so defined satisfy the rules of a system of maps $(f_\alpha )_\alpha $. To see this suppose that $\psi : [k'] \to [k]$, $\alpha : [k] \to [n + 1]$ with $k, k' \leq n$. Set $\alpha ' = \alpha \circ \psi $. Choose $i$ not in the image of $\alpha $. Then clearly $i$ is not in the image of $\alpha '$ also. Write $\alpha = \delta ^{n + 1}_ i \circ \phi $ (we cannot use the letter $\psi $ here because we've already used it). Then obviously $\alpha ' = \delta ^{n + 1}_ i \circ \phi \circ \psi $. By construction above we then have

as desired. We leave to the reader the pleasant task of verifying that our constructions are mutually inverse bijections, and are functorial in $T$. $\square$

Proof. By Lemma 14.19.3 there are identifications

for $0 \leq i \leq n$. By Lemma 14.19.5 we have

Thus we may define for any $\varphi : [i] \to [j]$ with $i, j \leq n + 1$ the corresponding map $\tilde U(\varphi ) : \tilde U_ j \to \tilde U_ i$ exactly as in Lemma 14.19.2. This defines an $(n + 1)$-truncated simplicial object $\tilde U$ with $\text{sk}_ n \tilde U = U$.

To see the adjointness we argue as follows. Given any element $\gamma : \text{sk}_ n V \to U$ of the right hand side of the formula consider the morphisms $f_ i = \gamma _ n \circ d^{n + 1}_ i : V_{n + 1} \to V_ n \to U_ n$. These clearly satisfy the relations $d^ n_{j - 1} \circ f_ i = d^ n_ i \circ f_ j$ and hence define a unique morphism $V_{n + 1} \to U_{n + 1}$ by our choice of $U_{n + 1}$. Conversely, given a morphism $\gamma ' : V \to \tilde U$ of the left hand side we can simply restrict to $\Delta _{\leq n}$ to get an element of the right hand side. We leave it to the reader to show these are mutually inverse constructions. $\square$

Proof. The existence in (1) follows from Lemma 14.19.2 above. Parts (2) and (3) follow from the discussion in Remark 14.19.9. After this (4) is obvious. $\square$

Proof. Let $W$ be a simplicial object. We have

The lemma follows. $\square$

Proof. Omitted, but very similar to the proof of Lemma 14.19.12 above. $\square$

Proof. The statement means that if $U$ is a simplicial object of $\mathcal{C}/X$ which we can think of as a simplicial object of $\mathcal{C}$ with a morphism towards the constant simplicial object $X$, then $\text{cosk}_ k U$ computed in $\mathcal{C}/X$ is the same as computed in $\mathcal{C}$. This follows for example from Categories, Lemma 4.16.2 because the categories $(\Delta /[n])_{\leq k}$ for $k \geq 1$ and $n \geq k + 1$ used in Lemma 14.19.2 are connected. Observe that we do not need the categories for $n \leq k$ by Lemma 14.19.3 or Lemma 14.19.4. $\square$

Proof. Consider a simplicial set $U$ and a morphism $f : U \to \Delta [n]$. This is a rule that associates to each $u \in U_ i$ a map $f_ u : [i] \to [n]$ in $\Delta $. Furthermore, these maps should have the property that $f_ u \circ \varphi = f_{U(\varphi )(u)}$ for any $\varphi : [j] \to [i]$. Denote $\epsilon ^ i_ j : [0] \to [i]$ the map which maps $0$ to $j$. Denote $F : U_0 \to [n]$ the map $u \mapsto f_ u(0)$. Then we see that

for all $0 \leq j \leq i$ and $u \in U_ i$. In particular, if we know the function $F$ then we know the maps $f_ u$ for all $u\in U_ i$ all $i$. Conversely, given a map $F : U_0 \to [n]$, we can set for any $i$, and any $u \in U_ i$ and any $0 \leq j \leq i$

This does not in general define a morphism $f$ of simplicial sets as above. Namely, the condition is that all the maps $f_ u$ are nondecreasing. This clearly is equivalent to the condition that $F(\epsilon ^ i_ j(u)) \leq F(\epsilon ^ i_{j'}(u))$ whenever $0 \leq j \leq j' \leq i$ and $u \in U_ i$. But in this case the morphisms

both factor through the map $\epsilon ^ i_{j, j'} : [1] \to [i]$ defined by the rules $0 \mapsto j$, $1 \mapsto j'$. In other words, it is enough to check the inequalities for $i = 1$ and $u \in X_1$. In other words, we have

as desired. $\square$