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Tag 017O

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 \begin{equation} \tag{14.18.1.1} \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 the 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) no two nondegenerate simplices of $U$ are mapped to the same simplex of $V$. 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 $$ U'_m = \bigcup\nolimits_{\varphi : [m] \to [i], ~i\leq n} \mathop{\rm Im}(U(\varphi)) $$ 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 $$ N(U_m) = \bigcap\nolimits_{i = 0}^{m - 1} \mathop{\rm Ker}(d^m_i). $$ 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 $x'' \in \mathop{\rm Im}(s_i)$ 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 $x_i \in \mathop{\rm Ker}(d_0) \cap \mathop{\rm Ker}(d_1) \cap \ldots \cap \mathop{\rm Ker}(d_i)$ and $z_i \in \mathop{\rm Ker}(d_0) \cap \ldots \cap \mathop{\rm Ker}(d_{i - 1})$ for $i = n, \ldots, 0$. Here and in the following an empty intersection of kernels indicates the whole space; i.e., the notation $z_0 \in \mathop{\rm Ker}(d_0) \cap \ldots \cap \mathop{\rm Ker}(d_{i - 1})$ 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 $z_i \in \mathop{\rm Ker}(d_0) \cap \ldots \cap \mathop{\rm Ker}(d_{i - 1})$. We can reformulate this as saying that we have found a direct sum decomposition $$ U_{n + 1} = N(U_{n + 1}) \oplus \bigoplus\nolimits_{i = 0}^{i = n} s_i\Big(\mathop{\rm Ker}(d_0) \cap \ldots \cap \mathop{\rm Ker}(d_{i - 1})\Big) $$ with the property that $$ \mathop{\rm Ker}(d_0) \cap \ldots \cap \mathop{\rm Ker}(d_j) = N(U_{n + 1}) \oplus \bigoplus\nolimits_{i = j + 1}^{i = n} s_i\Big(\mathop{\rm Ker}(d_n) \cap \ldots \cap \mathop{\rm Ker}(d_{i - 1})\Big) $$ 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 $z'_{i, j} \in \mathop{\rm Ker}(d_0) \cap \ldots \cap \mathop{\rm Ker}(d_{j - 1})$. 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 $$ N(U_m) = \bigcap\nolimits_{i = 0}^{m - 1} \mathop{\rm Ker}(d^m_i). $$ 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 $\mathop{\rm Mor}\nolimits_\mathcal{A}(A, U)$. Each of these are canonically split by Lemma 14.18.5. Moreover, $$ N(\mathop{\rm Mor}\nolimits_\mathcal{A}(A, U_m)) = \bigcap\nolimits_{i = 0}^{m - 1} \mathop{\rm Ker}(d^m_i) = \mathop{\rm Mor}\nolimits_\mathcal{A}(A, N(U_m)). $$ Hence we see that the morphism (14.18.1.1) becomes an isomorphism after applying the functor $\mathop{\rm Mor}\nolimits_\mathcal{A}(A, -)$ 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 $$ U'_m = \sum\nolimits_{\varphi : [m] \to [i], ~i\leq n} \mathop{\rm Im}(U(\varphi)) $$ 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 $\mathop{\rm Im}(U(\varphi))$ 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 $$ \sum\nolimits_{\varphi : [m] \to [m']\text{ surjective}, ~m' < m} \mathop{\rm Im}(U(\varphi)|_{N(U_{m'})}). $$ 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 $$ \sum\nolimits_{\varphi : [m] \to [m']\text{ surjective}, ~m' < m} \mathop{\rm Im}(U(\varphi)). $$ 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. $\square$

    The code snippet corresponding to this tag is a part of the file simplicial.tex and is located in lines 1742–2129 (see updates for more information).

    \section{Splitting simplicial objects}
    \label{section-splitting}
    
    \noindent
    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.)
    
    \begin{definition}
    \label{definition-split}
    Let $\mathcal{C}$ be a category which admits finite nonempty coproducts.
    We say a simplicial object $U$ of $\mathcal{C}$ is {\it split}
    if there exist subobjects $N(U_m)$ of $U_m$, $m \geq 0$
    with the property that
    \begin{equation}
    \label{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 {\it split}
    if there exist subobjects $N(U_m)$ of $U_m$, $r \geq m \geq 0$
    with the property that (\ref{equation-splitting})
    is an isomorphism for $r \geq n \geq 0$.
    \end{definition}
    
    \noindent
    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.
    
    \begin{lemma}
    \label{lemma-splitting-simplicial-sets}
    Let $U$ be a simplicial set. Then $U$ has a unique splitting
    with $N(U_m)$ equal to the set of nondegenerate $m$-simplices.
    \end{lemma}
    
    \begin{proof}
    From the definition it follows immediately, that if there is a
    splitting then $N(U_m)$ has the 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 $\psi = \varphi$.
    \end{proof}
    
    \noindent
    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 \ref{lemma-splitting-simplicial-sets}
    is not functorial. Here is a case where it is functorial.
    
    \begin{lemma}
    \label{lemma-injective-map-simplicial-sets}
    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)
    no two nondegenerate simplices of $U$ are mapped
    to the same simplex of $V$.
    Then $f_n$ is injective for all $n$.
    Same holds with ``injective'' replaced by
    ``surjective'' or ``bijective''.
    \end{lemma}
    
    \begin{proof}
    Under hypothesis (a) we see that the map $f$ preserves
    the disjoint union decompositions of the splitting
    of Lemma \ref{lemma-splitting-simplicial-sets}, 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).
    \end{proof}
    
    \begin{lemma}
    \label{lemma-simplicial-set-n-skel-sub}
    Let $U$ be a simplicial set.
    Let $n \geq 0$ be an integer.
    The rule
    $$
    U'_m = \bigcup\nolimits_{\varphi : [m] \to [i], \ i\leq n} \Im(U(\varphi))
    $$
    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 $m > n$.
    \end{lemma}
    
    \begin{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.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-splitting-simplicial-groups}
    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
    $$
    N(U_m) = \bigcap\nolimits_{i = 0}^{m - 1} \Ker(d^m_i).
    $$
    Moreover, this splitting is functorial on the category
    of simplicial abelian groups.
    \end{lemma}
    
    \begin{proof}
    By induction on $n$ we will show that the choice of $N(U_m)$
    in the lemma guarantees that (\ref{equation-splitting}) 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 \ref{remark-relations}.
    
    \medskip\noindent
    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 $x'' \in \Im(s_i)$
    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.
    
    \medskip\noindent
    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
    $x_i \in
    \Ker(d_0) \cap
    \Ker(d_1) \cap
    \ldots \cap
    \Ker(d_i)$
    and
    $z_i \in
    \Ker(d_0) \cap
    \ldots \cap
    \Ker(d_{i - 1})$
    for $i = n, \ldots, 0$.
    Here and in the following
    an empty intersection of kernels indicates
    the whole space; i.e.,
    the notation
    $z_0 \in \Ker(d_0) \cap
    \ldots \cap
    \Ker(d_{i - 1})$
    when $i = 0$ means $z_0 \in U_n$ with no restriction.
    
    \medskip\noindent
    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 \ref{remark-relations}.
    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.
    
    \medskip\noindent
    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
    $z_i \in \Ker(d_0) \cap \ldots \cap \Ker(d_{i - 1})$.
    We can reformulate this as saying that we have found a direct
    sum decomposition
    $$
    U_{n + 1}
    =
    N(U_{n + 1})
    \oplus
    \bigoplus\nolimits_{i = 0}^{i = n}
    s_i\Big(\Ker(d_0) \cap \ldots \cap \Ker(d_{i - 1})\Big)
    $$
    with the property that
    $$
    \Ker(d_0) \cap \ldots \cap \Ker(d_j)
    =
    N(U_{n + 1}) \oplus
    \bigoplus\nolimits_{i = j + 1}^{i = n}
    s_i\Big(\Ker(d_n) \cap \ldots \cap \Ker(d_{i - 1})\Big)
    $$
    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
    $z'_{i, j} \in \Ker(d_0) \cap \ldots \cap \Ker(d_{j - 1})$.
    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 $0 \leq i_1 \leq i_2 \leq \ldots \leq i_k \leq n - k + 1$.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-splitting-abelian-category}
    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
    $$
    N(U_m) = \bigcap\nolimits_{i = 0}^{m - 1} \Ker(d^m_i).
    $$
    Moreover, this splitting is functorial on the category of
    simplicial objects of $\mathcal{A}$.
    \end{lemma}
    
    \begin{proof}
    For any object $A$ of $\mathcal{A}$ we obtain
    a simplicial abelian group $\Mor_\mathcal{A}(A, U)$.
    Each of these are canonically split by Lemma
    \ref{lemma-splitting-simplicial-groups}. Moreover,
    $$
    N(\Mor_\mathcal{A}(A, U_m)) =
    \bigcap\nolimits_{i = 0}^{m - 1} \Ker(d^m_i) =
    \Mor_\mathcal{A}(A, N(U_m)).
    $$
    Hence we see that the morphism (\ref{equation-splitting})
    becomes an isomorphism after applying the functor
    $\Mor_\mathcal{A}(A, -)$ for any object of $\mathcal{A}$.
    Hence it is an isomorphism by the Yoneda lemma.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-injective-map-simplicial-abelian}
    \begin{slogan}
    The Dold-Kan normalization functor reflects
    injectivity, surjectivity, and isomorphy.
    \end{slogan}
    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''.
    \end{lemma}
    
    \begin{proof}
    This is clear from Lemma \ref{lemma-splitting-abelian-category}
    and the definition of a splitting.
    \end{proof}
    
    
    \begin{lemma}
    \label{lemma-N-d-in-N}
    Let $\mathcal{A}$ be an abelian category.
    Let $U$ be a simplicial object in $\mathcal{A}$.
    Let $N(U_m)$ as in Lemma \ref{lemma-splitting-abelian-category} above.
    Then $d^m_m(N(U_m)) \subset N(U_{m - 1})$.
    \end{lemma}
    
    \begin{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 \ref{remark-relations}.
    The result follows.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-simplicial-abelian-n-skel-sub}
    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
    $$
    U'_m = \sum\nolimits_{\varphi : [m] \to [i], \ i\leq n} \Im(U(\varphi))
    $$
    defines a sub simplicial object $U' \subset U$ with $U'_i = U_i$
    for $i \leq n$.
    Moreover, $N(U'_m) = 0$ for all $m > n$.
    \end{lemma}
    
    \begin{proof}
    Pick $m$, $i \leq n$ and some $\varphi : [m] \to [i]$.
    The image under $U(\psi)$ of $\Im(U(\varphi))$
    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 \ref{lemma-splitting-abelian-category},
    it suffices to show that $U'_m$ is contained in the sum
    $$
    \sum\nolimits_{\varphi : [m] \to [m']\text{ surjective}, \ m' < m}
    \Im(U(\varphi)|_{N(U_{m'})}).
    $$
    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
    $$
    \sum\nolimits_{\varphi : [m] \to [m']\text{ surjective}, \ m' < m}
    \Im(U(\varphi)).
    $$
    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.
    \end{proof}

    Comments (3)

    Comment #1851 by xai on March 1, 2016 a 6:06 am UTC

    typo in the second line: by abuse of notation

    Comment #1890 by Johan (site) on April 1, 2016 a 11:18 pm UTC

    Thanks, fixed here.

    Comment #2643 by Sebastian Ørsted on July 11, 2017 a 11:55 am UTC

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

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