The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

14.29 More homotopies in abelian categories

Let $\mathcal{A}$ be an abelian category. In this section we show that a homotopy between morphisms in $\text{Ch}_{\geq 0}(\mathcal{A})$ always comes from a morphism $U \times \Delta [1] \to V$ in the category of simplicial objects. In some sense this will provide a converse to Lemma 14.27.1. We first develop some material on homotopies between morphisms of chain complexes.

Lemma 14.29.1. Let $\mathcal{A}$ be an abelian category. Let $A$ be a chain complex. Consider the covariant functor

\[ B \longmapsto \{ (a, b, h) \mid a, b : A \to B\text{ and }h\text{ a homotopy between }a, b \} \]

There exists a chain complex $\diamond A$ such that $\mathop{Mor}\nolimits _{\text{Ch}(\mathcal{A})}(\diamond A, -)$ is isomorphic to the displayed functor. The construction $A \mapsto \diamond A$ is functorial.

Proof. We set $\diamond A_ n = A_ n \oplus A_ n \oplus A_{n - 1}$, and we define $d_{\diamond A, n}$ by the matrix

\[ d_{\diamond A, n} = \left( \begin{matrix} d_{A, n} & 0 & \text{id}_{A_{n - 1}} \\ 0 & d_{A, n} & -\text{id}_{A_{n - 1}} \\ 0 & 0 & -d_{A, n - 1} \end{matrix} \right) : A_ n \oplus A_ n \oplus A_{n - 1} \to A_{n - 1} \oplus A_{n - 1} \oplus A_{n - 2} \]

If $\mathcal{A}$ is the category of abelian groups, and $(x, y, z) \in A_ n \oplus A_ n \oplus A_{n - 1}$ then $d_{\diamond A, n}(x, y, z) = (d_ n(x) + z, d_ n(y) - z, -d_{n - 1}(z))$. It is easy to verify that $d^2 = 0$. Clearly, there are two maps $\diamond a, \diamond b : A \to \diamond A$ (first summand and second summand), and a map $\diamond A \to A[-1]$ which give a short exact sequence

\[ 0 \to A \oplus A \to \diamond A \to A[-1] \to 0 \]

which is termwise split. Moreover, there is a sequence of maps $\diamond h_ n : A_ n \to \diamond A_{n + 1}$, namely the identity from $A_ n$ to the summand $A_ n$ of $\diamond A_{n + 1}$, such that $\diamond h$ is a homotopy between $\diamond a$ and $\diamond b$.

We conclude that any morphism $f : \diamond A \to B$ gives rise to a triple $(a, b, h)$ by setting $a = f \circ \diamond a$, $b = f \circ \diamond b$ and $h_ n = f_{n + 1} \circ \diamond h_ n$. Conversely, given a triple $(a, b, h)$ we get a morphism $f : \diamond A \to B$ by taking

\[ f_ n = (a_ n, b_ n, h_{n - 1}). \]

To see that this is a morphism of chain complexes you have to do a calculation. We only do this in case $\mathcal{A}$ is the category of abelian groups: Say $(x, y, z) \in \diamond A_ n = A_ n \oplus A_ n \oplus A_{n - 1}$. Then

\begin{eqnarray*} f_{n - 1}(d_ n(x, y, z)) & = & f_{n - 1}(d_ n(x) + z, d_ n(y) - z, -d_{n - 1}(z)) \\ & = & a_ n(d_ n(x)) + a_ n(z) + b_ n(d_ n(y)) - b_ n(z) - h_{n - 2}(d_{n - 1}(z)) \end{eqnarray*}

and

\begin{eqnarray*} d_ n(f_ n(x, y, z) & = & d_ n(a_ n(x) + b_ n(y) + h_{n - 1}(z)) \\ & = & d_ n(a_ n(x)) + d_ n(b_ n(y)) + d_ n(h_{n - 1}(z)) \end{eqnarray*}

which are the same by definition of a homotopy. $\square$

Note that the extension

\[ 0 \to A \oplus A \to \diamond A \to A[-1] \to 0 \]

comes with sections of the morphisms $\diamond A_ n \to A[-1]_ n$ with the property that the associated morphism $\delta : A[-1] \to (A \oplus A)[-1]$, see Homology, Lemma 12.13.4 equals the morphism $(1, -1) : A[-1] \to A[-1] \oplus A[-1]$.

Lemma 14.29.2. Let $\mathcal{A}$ be an abelian category. Let

\[ 0 \to A \oplus A \to B \to C \to 0 \]

be a short exact sequence of chain complexes of $\mathcal{A}$. Suppose given in addition morphisms $s_ n : C_ n \to B_ n$ splitting the associated short exact sequence in degree $n$. Let $\delta (s) : C \to (A \oplus A)[-1] = A[-1] \oplus A[-1]$ be the associated morphism of complexes, see Homology, Lemma 12.13.4. If $\delta (s)$ factors through the morphism $(1, -1) : A[-1] \to A[-1] \oplus A[-1]$, then there is a unique morphism $B \to \diamond A$ fitting into a commutative diagram

\[ \xymatrix{ 0 \ar[r] & A \oplus A \ar[d] \ar[r] & B \ar[r] \ar[d] & C \ar[d] \ar[r] & 0 \\ 0 \ar[r] & A \oplus A \ar[r] & \diamond A \ar[r] & A[-1] \ar[r] & 0 } \]

where the vertical maps are compatible with the splittings $s_ n$ and the splittings of $\diamond A_ n \to A[-1]_ n$ as well.

Proof. Denote $(p_ n, q_ n) : B_ n \to A_ n \oplus A_ n$ the morphism $\pi _ n$ of Homology, Lemma 12.13.4. Also write $(a, b) : A \oplus A \to B$, and $r : B \to C$ for the maps in the short exact sequence. Write the factorization of $\delta (s)$ as $\delta (s) = (1, -1) \circ f$. This means that $p_{n - 1} \circ d_{B, n} \circ s_ n = f_ n$, and $q_{n - 1} \circ d_{B, n} \circ s_ n = - f_ n$, and Set $B_ n \to \diamond A_ n = A_ n \oplus A_ n \oplus A_{n - 1}$ equal to $(p_ n, q_ n, f_ n \circ r_ n)$.

Now we have to check that this actually defines a morphism of complexes. We will only do this in the case of abelian groups. Pick $x \in B_ n$. Then $x = a_ n(x_1) + b_ n(x_2) + s_ n(x_3)$ and it suffices to show that our definition commutes with differential for each term separately. For the term $a_ n(x_1)$ we have $(p_ n, q_ n, f_ n \circ r_ n)(a_ n(x_1)) = (x_1, 0, 0)$ and the result is obvious. Similarly for the term $b_ n(x_2)$. For the term $s_ n(x_3)$ we have

\begin{eqnarray*} (p_ n, q_ n, f_ n \circ r_ n)(d_ n(s_ n(x_3))) & = & (p_ n, q_ n, f_ n \circ r_ n)( \\ & & \ \ \ \ \ a_ n(f_ n(x_3)) - b_ n(f_ n(x_3)) + s_ n(d_ n(x_3))) \\ & = & (f_ n(x_3), -f_ n(x_3), f_ n(d_ n(x_3))) \end{eqnarray*}

by definition of $f_ n$. And

\begin{eqnarray*} d_ n(p_ n, q_ n, f_ n \circ r_ n)(s_ n(x_3)) & = & d_ n(0, 0, f_ n(x_3)) \\ & = & (f_ n(x_3), - f_ n(x_3), d_{A[-1], n}(f_ n(x_3))) \end{eqnarray*}

The result follows as $f$ is a morphism of complexes. $\square$

Lemma 14.29.3. Let $\mathcal{A}$ be an abelian category. Let $U$, $V$ be simplicial objects of $\mathcal{A}$. Let $a, b : U \to V$ be a pair of morphisms. Assume the corresponding maps of chain complexes $N(a), N(b) : N(U) \to N(V)$ are homotopic by a homotopy $\{ N_ n : N(U)_ n \to N(V)_{n + 1}\} $. Then $a, b$ are homotopic in the sense of Definition 14.26.1. Moreover, one can choose the homotopy $h : U \times \Delta [1] \to V$ such that $N_ n = N(h)_ n$ where $N(h)$ is the homotopy coming from $h$ as in Section 14.27.

Proof. Let $(\diamond N(U), \diamond a, \diamond b, \diamond h)$ be as in Lemma 14.29.1 and its proof. By that lemma there exists a morphism $\diamond N(U) \to N(V)$ representing the triple $(N(a), N(b), \{ N_ n\} )$. We will show there exists a morphism $\psi : N(U \times \Delta [1]) \to \diamond {N(U)}$ such that $\diamond a = \psi \circ N(e_0)$, and $\diamond b = \psi \circ N(e_1)$. Moreover, we will show that the homotopy between $N(e_0), N(e_1) : N(U) \to N(U \times \Delta [1])$ coming from (14.27.0.1) and Lemma 14.27.1 with $h = \text{id}_{U \times \Delta [1]}$ is mapped via $\psi $ to the canonical homotopy $\diamond h$ between the two maps $\diamond a, \diamond b : N(U) \to \diamond {N(U)}$. Certainly this will imply the lemma.

Note that $N : \text{Simp}(\mathcal{A}) \to \text{Ch}_{\geq 0}(\mathcal{A})$ as a functor is a direct summand of the functor $N : \text{Simp}(\mathcal{A}) \to \text{Ch}_{\geq 0}(\mathcal{A})$. Also, the functor $\diamond $ is compatible with direct sums. Thus it suffices instead to construct a morphism $\Psi : s(U \times \Delta [1]) \to \diamond {s(U)}$ with the corresponding properties. This is what we do below.

By Definition 14.26.1 the morphisms $e_0 : U \to U \times \Delta [1]$ and $e_1 : U \to U \times \Delta [1]$ are homotopic with homotopy $\text{id}_{U \times \Delta [1]}$. By Lemma 14.27.1 we get an explicit homotopy $\{ h_ n : s(U)_ n \to s(U \times \Delta [1])_{n + 1}\} $ between the morphisms of chain complexes $s(e_0) : s(U) \to s(U \times \Delta [1])$ and $s(e_1) : s(U) \to s(U \times \Delta [1])$. By Lemma 14.29.2 above we get a corresponding morphism

\[ \Phi : \diamond {s(U)} \to s(U \times \Delta [1]) \]

According to the construction, $\Phi _ n$ restricted to the summand $s(U)[-1]_ n = s(U)_{n - 1}$ of $\diamond {s(U)}_ n$ is equal to $h_{n - 1}$. And

\[ h_{n - 1} = \sum \nolimits _{i = 0}^{n - 1} (-1)^{i + 1} s^ n_ i \cdot \alpha ^ n_{i + 1} : U_{n - 1} \to \bigoplus \nolimits _ j U_ n \cdot \alpha ^ n_ j. \]

with obvious notation.

On the other hand, the morphisms $e_ i : U \to U \times \Delta [1]$ induce a morphism $(e_0, e_1) : U \oplus U \to U \times \Delta [1]$. Denote $W$ the cokernel. Note that, if we write $(U \times \Delta [1])_ n = \bigoplus _{\alpha : [n] \to [1]} U_ n \cdot \alpha $, then we may identify $W_ n = \bigoplus _{i = 1}^ n U_ n \cdot \alpha ^ n_ i$ with $\alpha ^ n_ i$ as in Section 14.26. We have a commutative diagram

\[ \xymatrix{ 0 \ar[r] & U \oplus U \ar[rd]_{(1, 1)} \ar[r] & U \times \Delta [1] \ar[d]^\pi \ar[r] & W \ar[r] & 0 \\ & & U & & } \]

This implies we have a similar commutative diagram after applying the functor $s$. Next, we choose the splittings $\sigma _ n : s(W)_ n \to s(U \times \Delta [1])_ n$ by mapping the summand $U_ n \cdot \alpha ^ n_ i \subset W_ n$ via $(-1, 1)$ to the summands $U_ n \cdot \alpha ^ n_0 \oplus U_ n \cdot \alpha ^ n_ i \subset (U \times \Delta [1])_ n$. Note that $s(\pi )_ n \circ \sigma _ n = 0$. It follows that $(1, 1) \circ \delta (\sigma )_ n = 0$. Hence $\delta (\sigma )$ factors as in Lemma 14.29.2. By that lemma we obtain a canonical morphism $\Psi : s(U \times \Delta [1]) \to \diamond {s(U)}$.

To compute $\Psi $ we first compute the morphism $\delta (\sigma ) : s(W) \to s(U)[-1] \oplus s(U)[-1]$. According to Homology, Lemma 12.13.4 and its proof, to do this we have compute

\[ d_{s(U \times \delta [1]), n} \circ \sigma _ n - \sigma _{n - 1} \circ d_{s(W), n} \]

and write it as a morphism into $U_{n - 1} \cdot \alpha ^{n - 1}_0 \oplus U_{n - 1} \cdot \alpha ^{n - 1}_ n$. We only do this in case $\mathcal{A}$ is the category of abelian groups. We use the short hand notation $x_{\alpha }$ for $x \in U_ n$ to denote the element $x$ in the summand $U_ n \cdot \alpha $ of $(U \times \Delta [1])_ n$. Recall that

\[ d_{s(U \times \delta [1]), n} = \sum \nolimits _{i = 0}^ n (-1)^ i d^ n_ i \]

where $d^ n_ i$ maps the summand $U_ n \cdot \alpha $ to the summand $U_{n - 1} \cdot (\alpha \circ \delta ^ n_ i)$ via the morphism $d^ n_ i$ of the simplicial object $U$. In terms of the notation above this means

\[ d_{s(U \times \delta [1]), n}(x_\alpha ) = \sum \nolimits _{i = 0}^ n (-1)^ i (d^ n_ i(x))_{\alpha \circ \delta ^ n_ i} \]

Starting with $x_\alpha \in W_ n$, in other words $\alpha = \alpha ^ n_ j$ for some $j \in \{ 1, \ldots , n\} $, we see that $\sigma _ n(x_\alpha ) = x_\alpha - x_{\alpha ^ n_0}$ and hence

\[ (d_{s(U \times \delta [1]), n} \circ \sigma _ n)(x_\alpha ) = \sum \nolimits _{i = 0}^ n (-1)^ i (d^ n_ i(x))_{\alpha \circ \delta ^ n_ i} - \sum \nolimits _{i = 0}^ n (-1)^ i (d^ n_ i(x))_{\alpha ^ n_0 \circ \delta ^ n_ i} \]

To compute $d_{s(W), n}(x_\alpha )$, we have to omit all terms where $\alpha \circ \delta ^ n_ i = \alpha ^{n - 1}_0, \alpha ^{n - 1}_ n$. Hence we get

\[ \begin{matrix} (\sigma _{n - 1} \circ d_{s(W), n})(x_\alpha ) = \\ \sum \nolimits _{i = 0, \ldots , n\text{ and } \alpha \circ \delta ^ n_ i \not= \alpha ^{n - 1}_0\text{ or }\alpha ^{n - 1}_ n} \Big((-1)^ i (d^ n_ i(x))_{\alpha \circ \delta ^ n_ i} - (-1)^ i (d^ n_ i(x))_{\alpha ^{n - 1}_0} \Big) \end{matrix} \]

Clearly the difference of the two terms is the sum

\[ \sum \nolimits _{i = 0, \ldots , n\text{ and } \alpha \circ \delta ^ n_ i = \alpha ^{n - 1}_0\text{ or }\alpha ^{n - 1}_ n} \Big((-1)^ i (d^ n_ i(x))_{\alpha \circ \delta ^ n_ i} - (-1)^ i (d^ n_ i(x))_{\alpha ^{n - 1}_0} \Big) \]

Of course, if $\alpha \circ \delta ^ n_ i = \alpha ^{n - 1}_0$ then the term drops out. Recall that $\alpha = \alpha ^ n_ j$ for some $j \in \{ 1, \ldots , n\} $. The only way $\alpha ^ n_ j \circ \delta ^ n_ i = \alpha ^{n - 1}_ n$ is if $j = n$ and $i = n$. Thus we actually get $0$ unless $j = n$ and in that case we get $(-1)^ n(d^ n_ n(x))_{\alpha ^{n - 1}_ n} - (-1)^ n(d^ n_ n(x))_{\alpha ^{n - 1}_0}$. In other words, we conclude the morphism

\[ \delta (\sigma )_ n : W_ n \to (s(U)[-1] \oplus s(U)[-1])_ n = U_{n - 1} \oplus U_{n - 1} \]

is zero on all summands except $U_ n \cdot \alpha ^ n_ n$ and on that summand it is equal to $((-1)^ nd^ n_ n, -(-1)^ nd^ n_ n)$. (Namely, the first summand of the two corresponds to the factor with $\alpha ^{n - 1}_ n$ because that is the map $[n - 1] \to [1]$ which maps everybody to $0$, and hence corresponds to $e_0$.)

We obtain a canonical diagram

\[ \xymatrix{ 0 \ar[r] & s(U) \oplus s(U) \ar[r] \ar[d] & \diamond {s(U)} \ar[r] \ar[d]^{\Phi }& s(U)[-1] \ar[r] \ar[d] & 0 \\ 0 \ar[r] & s(U) \oplus s(U) \ar[r] \ar[d] & s(U \times \Delta [1]) \ar[r] \ar[d]^\Psi & s(W) \ar[r] \ar[d] & 0 \\ 0 \ar[r] & s(U) \oplus s(U) \ar[r] & \diamond {s(U)} \ar[r] & s(U)[-1] \ar[r] & 0 } \]

We claim that $\Phi \circ \Psi $ is the identity. To see this it is enough to prove that the composition of $\Phi $ and $\delta (\sigma )$ as a map $s(U)[-1] \to s(W) \to s(U)[-1] \oplus s(U)[-1]$ is the identity in the first factor and minus identity in the second. By the computations above it is $((-1)^ nd^ n_0, -(-1)^ nd^ n_0) \circ (-1)^ n s^ n_ n = (1, -1)$ as desired. $\square$


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