## 14.18 Splitting simplicial objects

A subobject $N$ of an object $X$ of the category $\mathcal{C}$ is an object $N$ of $\mathcal{C}$ together with a monomorphism $N \to X$. Of course we say (by abuse of notation) that the subobjects $N$, $N'$ are equal if there exists an isomorphism $N \to N'$ compatible with the morphisms to $X$. The collection of subobjects forms a partially ordered set. (Because of our conventions on categories; not true for category of spaces up to homotopy for example.)

Definition 14.18.1. Let $\mathcal{C}$ be a category which admits finite nonempty coproducts. We say a simplicial object $U$ of $\mathcal{C}$ is split if there exist subobjects $N(U_ m)$ of $U_ m$, $m \geq 0$ with the property that

14.18.1.1
\begin{equation} \label{simplicial-equation-splitting} \coprod \nolimits _{\varphi : [n] \to [m]\text{ surjective}} N(U_ m) \longrightarrow U_ n \end{equation}

is an isomorphism for all $n \geq 0$. If $U$ is an $r$-truncated simplicial object of $\mathcal{C}$ then we say $U$ is split if there exist subobjects $N(U_ m)$ of $U_ m$, $r \geq m \geq 0$ with the property that (14.18.1.1) is an isomorphism for $r \geq n \geq 0$.

If this is the case, then $N(U_0) = U_0$. Next, we have $U_1 = U_0 \amalg N(U_1)$. Second we have

$U_2 = U_0 \amalg N(U_1) \amalg N(U_1) \amalg N(U_2).$

It turns out that in many categories $\mathcal{C}$ every simplicial object is split.

Lemma 14.18.2. Let $U$ be a simplicial set. Then $U$ has a unique splitting with $N(U_ m)$ equal to the set of nondegenerate $m$-simplices.

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 $\psi = \varphi$. $\square$

Of course it can happen that a map of simplicial sets maps a nondegenerate $n$-simplex to a degenerate $n$-simplex. Thus the splitting of Lemma 14.18.2 is not functorial. Here is a case where it is functorial.

Lemma 14.18.3. Let $f : U \to V$ be a morphism of simplicial sets. Suppose that (a) the image of every nondegenerate simplex of $U$ is a nondegenerate simplex of $V$ and (b) the restriction of $f$ to a map from the set of nondegenerate simplices of $U$ to the set of nondegenerate simplices of $V$ is injective. Then $f_ n$ is injective for all $n$. Same holds with “injective” replaced by “surjective” or “bijective”.

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

$\xymatrix{ \coprod \nolimits _{\varphi : [n] \to [m]\text{ surjective}} N(U_ m) \ar[r] \ar[d] & U_ n \ar[d] \\ \coprod \nolimits _{\varphi : [n] \to [m]\text{ surjective}} N(V_ m) \ar[r] & V_ n. }$

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

Lemma 14.18.4. Let $U$ be a simplicial set. Let $n \geq 0$ be an integer. The rule

$U'_ m = \bigcup \nolimits _{\varphi : [m] \to [i], \ i\leq n} \mathop{\mathrm{Im}}(U(\varphi ))$

defines a sub simplicial set $U' \subset U$ with $U'_ i = U_ i$ for $i \leq n$. Moreover, all $m$-simplices of $U'$ are degenerate for all $m > n$.

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. $\square$

Lemma 14.18.5. Let $U$ be a simplicial abelian group. Then $U$ has a splitting obtained by taking $N(U_0) = U_0$ and for $m \geq 1$ taking

$N(U_ m) = \bigcap \nolimits _{i = 0}^{m - 1} \mathop{\mathrm{Ker}}(d^ m_ i).$

Moreover, this splitting is functorial on the category of simplicial abelian groups.

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 $d_ n(x_1) = 0$. Continue like this to get

\begin{eqnarray*} x & = & x_0 + s_0(z_0), \\ x_0 & = & x_1 + s_1(z_1), \\ x_1 & = & x_2 + s_2(z_2), \\ \ldots & \ldots & \ldots \\ x_{n - 1} & = & x_ n + s_ n(z_ n) \end{eqnarray*}

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

$0 = d_ j(x_{i - 1}) = d_ j(x_ i) + d_ j(s_ i(z_ i)) = d_ j(x_ i) + s_{i - 1}(d_ j(z_ i)).$

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

$x = s_0(z_0) + s_1(z_1) + \ldots + s_ n(z_ n) + x_0$

with $x_0 \in N(U_{n + 1})$ 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

$U_{n + 1} = N(U_{n + 1}) \oplus \bigoplus \nolimits _{i = 0}^{i = n} s_ i\Big(\mathop{\mathrm{Ker}}(d_0) \cap \ldots \cap \mathop{\mathrm{Ker}}(d_{i - 1})\Big)$

with the property that

$\mathop{\mathrm{Ker}}(d_0) \cap \ldots \cap \mathop{\mathrm{Ker}}(d_ j) = N(U_{n + 1}) \oplus \bigoplus \nolimits _{i = j + 1}^{i = n} s_ i\Big(\mathop{\mathrm{Ker}}(d_ n) \cap \ldots \cap \mathop{\mathrm{Ker}}(d_{i - 1})\Big)$

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

$z_ i = s_ i(z'_{i, i}) + \ldots + s_{n - 1}(z'_{i, n - 1}) + z_{i, 0}$

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

$x = x_0 + s_0(z_{0, 0}) + s_1(z_{1, 0}) + \ldots + s_ n(z_{n, 0}) + \sum \nolimits _{0 \leq i \leq j \leq n - 1} s_ i(s_ j(z'_{i, j})).$

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

$s_{i_1} s_{i_2} \ldots s_{i_ k} (z)$

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

$\sigma ^{n - k}_{i_ k} \ldots \sigma ^{n - 1}_{i_2} \sigma ^ n_{i_1}$

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

Lemma 14.18.6. Let $\mathcal{A}$ be an abelian category. Let $U$ be a simplicial object in $\mathcal{A}$. Then $U$ has a splitting obtained by taking $N(U_0) = U_0$ and for $m \geq 1$ taking

$N(U_ m) = \bigcap \nolimits _{i = 0}^{m - 1} \mathop{\mathrm{Ker}}(d^ m_ i).$

Moreover, this splitting is functorial on the category of simplicial objects of $\mathcal{A}$.

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,

$N(\mathop{\mathrm{Mor}}\nolimits _\mathcal {A}(A, U_ m)) = \bigcap \nolimits _{i = 0}^{m - 1} \mathop{\mathrm{Ker}}(d^ m_ i) = \mathop{\mathrm{Mor}}\nolimits _\mathcal {A}(A, N(U_ m)).$

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. $\square$

Lemma 14.18.7. Let $\mathcal{A}$ be an abelian category. Let $f : U \to V$ be a morphism of simplicial objects of $\mathcal{A}$. If the induced morphisms $N(f)_ i : N(U)_ i \to N(V)_ i$ are injective for all $i$, then $f_ i$ is injective for all $i$. Same holds with “injective” replaced with “surjective”, or “isomorphism”.

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

Lemma 14.18.8. Let $\mathcal{A}$ be an abelian category. Let $U$ be a simplicial object in $\mathcal{A}$. Let $N(U_ m)$ as in Lemma 14.18.6 above. Then $d^ m_ m(N(U_ m)) \subset N(U_{m - 1})$.

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. $\square$

Lemma 14.18.9. Let $\mathcal{A}$ be an abelian category. Let $U$ be a simplicial object of $\mathcal{A}$. Let $n \geq 0$ be an integer. The rule

$U'_ m = \sum \nolimits _{\varphi : [m] \to [i], \ i\leq n} \mathop{\mathrm{Im}}(U(\varphi ))$

defines a sub simplicial object $U' \subset U$ with $U'_ i = U_ i$ for $i \leq n$. Moreover, $N(U'_ m) = 0$ for all $m > n$.

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

$\sum \nolimits _{\varphi : [m] \to [m']\text{ surjective}, \ m' < m} \mathop{\mathrm{Im}}(U(\varphi )|_{N(U_{m'})}).$

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

$\sum \nolimits _{\varphi : [m] \to [m']\text{ surjective}, \ m' < m} \mathop{\mathrm{Im}}(U(\varphi )).$

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. $\square$

Comment #1851 by xai on

typo in the second line: by abuse of notation

Comment #2643 by Sebastian Ørsted on

It would probably be worth remarking that you just reproved Dold–Kan.

Comment #2663 by on

@#2643: Feel free to make a small edit of the LaTeX file and submit it to the email address of the Stacks project: stacks.project@gmail.com

Comment #7070 by Gleb on

A small typo: it should be "has to be" instead of "has the be".

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