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Tag 017U

Chapter 14: Simplicial Methods > Section 14.18: Splitting simplicial objects

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

    The code snippet corresponding to this tag is a part of the file simplicial.tex and is located in lines 1868–1878 (see updates for more information).

    \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}

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