The Stacks project

Proof. From the definition it follows immediately, that if there is a splitting then $N(U_ m)$ has to be the set of nondegenerate simplices. Let $x \in U_ n$. Suppose that there are surjections $\varphi : [n] \to [k]$ and $\psi : [n] \to [l]$ and nondegenerate simplices $y \in U_ k$, $z \in U_ l$ such that $x = U(\varphi )(y)$ and $x = U(\psi )(z)$. Choose a right inverse $\xi : [l] \to [n]$ of $\psi $, i.e., $\psi \circ \xi = \text{id}_{[l]}$. Then $z = U(\xi )(x)$. Hence $z = U(\xi )(x) = U(\varphi \circ \xi )(y)$. Since $z$ is nondegenerate we conclude that $\varphi \circ \xi : [l] \to [k]$ is surjective, and hence $l \geq k$. Similarly $k \geq l$. Hence we see that $\varphi \circ \xi : [l] \to [k]$ has to be the identity map for any choice of right inverse $\xi $ of $\psi $. This easily implies that $\psi = \varphi $. $\square$

Proof. Under hypothesis (a) we see that the map $f$ preserves the disjoint union decompositions of the splitting of Lemma 14.18.2, in other words that we get commutative diagrams

And then (b) clearly shows that the left vertical arrow is injective (resp. surjective, resp. bijective). $\square$

Proof. If $x \in U_ m$ and $x = U(\varphi )(y)$ for some $y \in U_ i$, $i \leq n$ and some $\varphi : [m] \to [i]$ then any image $U(\psi )(x)$ for any $\psi : [m'] \to [m]$ is equal to $U(\varphi \circ \psi )(y)$ and $\varphi \circ \psi : [m'] \to [i]$. Hence $U'$ is a simplicial set. By construction all simplices in dimension $n + 1$ and higher are degenerate. $\square$

Proof. By induction on $n$ we will show that the choice of $N(U_ m)$ in the lemma guarantees that (14.18.1.1) is an isomorphism for $m \leq n$. This is clear for $n = 0$. In the rest of this proof we are going to drop the superscripts from the maps $d_ i$ and $s_ i$ in order to improve readability. We will also repeatedly use the relations from Remark 14.3.3.

First we make a general remark. For $0 \leq i \leq m$ and $z \in U_ m$ we have $d_ i(s_ i(z)) = z$. Hence we can write any $x \in U_{m + 1}$ uniquely as $x = x' + x''$ with $d_ i(x') = 0$ and $x'' \in \mathop{\mathrm{Im}}(s_ i)$ by taking $x' = (x - s_ i(d_ i(x)))$ and $x'' = s_ i(d_ i(x))$. Moreover, the element $z \in U_ m$ such that $x'' = s_ i(z)$ is unique because $s_ i$ is injective.

Here is a procedure for decomposing any $x \in U_{n + 1}$. First, write $x = x_0 + s_0(z_0)$ with $d_0(x_0) = 0$. Next, write $x_0 = x_1 + s_1(z_1)$ with $d_1(x_1) = 0$. Continue like this to get

where $d_ i(x_ i) = 0$ for all $i = n, \ldots , 0$. By our general remark above all of the $x_ i$ and $z_ i$ are determined uniquely by $x$. We claim that $x_ i \in \mathop{\mathrm{Ker}}(d_0) \cap \mathop{\mathrm{Ker}}(d_1) \cap \ldots \cap \mathop{\mathrm{Ker}}(d_ i)$ and $z_ i \in \mathop{\mathrm{Ker}}(d_0) \cap \ldots \cap \mathop{\mathrm{Ker}}(d_{i - 1})$ for $i = n, \ldots , 0$. Here and in the following an empty intersection of kernels indicates the whole space; i.e., the notation $z_0 \in \mathop{\mathrm{Ker}}(d_0) \cap \ldots \cap \mathop{\mathrm{Ker}}(d_{i - 1})$ when $i = 0$ means $z_0 \in U_ n$ with no restriction.

We prove this by ascending induction on $i$. It is clear for $i = 0$ by construction of $x_0$ and $z_0$. Let us prove it for $0 < i \leq n$ assuming the result for $i - 1$. First of all we have $d_ i(x_ i) = 0$ by construction. So pick a $j$ with $0 \leq j < i$. We have $d_ j(x_{i - 1}) = 0$ by induction. Hence

The last equality by the relations of Remark 14.3.3. These relations also imply that $d_{i - 1}(d_ j(x_ i)) = d_ j(d_ i(x_ i)) = 0$ because $d_ i(x_ i)= 0$ by construction. Then the uniqueness in the general remark above shows the equality $0 = x' + x'' = d_ j(x_ i) + s_{i - 1}(d_ j(z_ i))$ can only hold if both terms are zero. We conclude that $d_ j(x_ i) = 0$ and by injectivity of $s_{i - 1}$ we also conclude that $d_ j(z_ i) = 0$. This proves the claim.

The claim implies we can uniquely write

with $x_ n \in N(U_{n + 1})$ and $z_ i \in \mathop{\mathrm{Ker}}(d_0) \cap \ldots \cap \mathop{\mathrm{Ker}}(d_{i - 1})$. We can reformulate this as saying that we have found a direct sum decomposition

with the property that

for $j = 0, \ldots , n$. The result follows from this statement as follows. Each of the $z_ i$ in the expression for $x$ can be written uniquely as

with $z_{i, 0} \in N(U_ n)$ and $z'_{i, j} \in \mathop{\mathrm{Ker}}(d_0) \cap \ldots \cap \mathop{\mathrm{Ker}}(d_{j - 1})$. The first few steps in the decomposition of $z_ i$ are zero because $z_ i$ already is in the kernel of $d_0, \ldots , d_ i$. This in turn uniquely gives

Continuing in this fashion we see that we in the end obtain a decomposition of $x$ as a sum of terms of the form

with $0 \leq i_1 \leq i_2 \leq \ldots \leq i_ k \leq n - k + 1$ and $z \in N(U_{n + 1 - k})$. This is exactly the required decomposition, because any surjective map $[n + 1] \to [n + 1 - k]$ can be uniquely expressed in the form

with $0 \leq i_1 \leq i_2 \leq \ldots \leq i_ k \leq n - k + 1$. $\square$

Proof. For any object $A$ of $\mathcal{A}$ we obtain a simplicial abelian group $\mathop{\mathrm{Mor}}\nolimits _\mathcal {A}(A, U)$. Each of these are canonically split by Lemma 14.18.5. Moreover,

Hence we see that the morphism (14.18.1.1) becomes an isomorphism after applying the functor $\mathop{\mathrm{Mor}}\nolimits _\mathcal {A}(A, -)$ for any object of $\mathcal{A}$. Hence it is an isomorphism by the Yoneda lemma. $\square$

Proof. This is clear from Lemma 14.18.6 and the definition of a splitting. $\square$

Proof. For $j = 0, \ldots , m - 2$ we have $d^{m - 1}_ j d^ m_ m = d^{m - 1}_{m - 1} d^ m_ j$ by the relations in Remark 14.3.3. The result follows. $\square$

Proof. Pick $m$, $i \leq n$ and some $\varphi : [m] \to [i]$. The image under $U(\psi )$ of $\mathop{\mathrm{Im}}(U(\varphi ))$ for any $\psi : [m'] \to [m]$ is equal to the image of $U(\varphi \circ \psi )$ and $\varphi \circ \psi : [m'] \to [i]$. Hence $U'$ is a simplicial object. Pick $m > n$. We have to show $N(U'_ m) = 0$. By definition of $N(U_ m)$ and $N(U'_ m)$ we have $N(U'_ m) = U'_ m \cap N(U_ m)$ (intersection of subobjects). Since $U$ is split by Lemma 14.18.6, it suffices to show that $U'_ m$ is contained in the sum

By the splitting each $U_{m'}$ is the sum of images of $N(U_{m''})$ via $U(\psi )$ for surjective maps $\psi : [m'] \to [m'']$. Hence the displayed sum above is the same as

Clearly $U'_ m$ is contained in this by the simple fact that any $\varphi : [m] \to [i]$, $i \leq n$ occurring in the definition of $U'_ m$ may be factored as $[m] \to [m'] \to [i]$ with $[m] \to [m']$ surjective and $m' < m$ as in the last displayed sum above. $\square$