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*}


\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 ( 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. $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 01A1. Beware of the difference between the letter 'O' and the digit '0'.