The Stacks project

Proof. This follows immediately from the functorial properties of the fibre product (Lemma 14.7.2) and the definitions. $\square$

Proof. Omitted. $\square$

Proof. Omitted. Hint: use that for a countable sequence of surjections of sets the inverse limit is nonempty. $\square$

Proof. Omitted. $\square$

Proof. The following proof is basically just a translation into English of the proof in the reference mentioned above. Using the terminology as explained in the introduction to this section, suppose $f : \Lambda _ k[n] \to X$ is a morphism from a horn. Set $x_ i = f(\sigma _ i) \in X_{n - 1}$ for $i = 0, \ldots , \hat k, \ldots , n$. This means that for $i < j$ we have $d_ i x_ j = d_{j - 1} x_ i$ whenever $i, j \not= k$. We have to find an $x \in X_ n$ such that $x_ i = d_ ix$ for $i = 0, \ldots , \hat k, \ldots , n$.

We first prove there exists a $u \in X_ n$ such that $d_ i u = x_ i$ for $i < k$. This is trivial for $k = 0$. If $k > 0$, one defines by induction an element $u^ r \in X_ n$ such that $d_ i u^ r = x_ i$ for $0 \leq i \leq r$. Start with $u^0 = s_0x_0$. If $r < k - 1$, we set

An easy calculation shows that $d_ iy^ r = 1$ (unit element of the group $X_{n - 1}$) for $i \leq r$ and $d_{r + 1}y^ r = (d_{r + 1}u^ r)^{-1}x_{r + 1}$. It follows that $d_ iu^{r + 1} = x_ i$ for $i \leq r + 1$. Finally, take $u = u^{k - 1}$ to get $u$ as promised.

Next we prove, by induction on the integer $r$, $0 \leq r \leq n - k$, there exists a $x^ r \in X_ n$ such that

Start with $x^0 = u$ for $r = 0$. Having defined $x^ r$ for $r \leq n - k - 1$ we set

A simple calculation, using the given relations, shows that $d_ iz^ r = 1$ for $i < k$ and $i > n - r$ and that $d_{n - r}(z^ r) = (d_{n - r}x^ r)^{-1}x_{n - r}$. It follows that $d_ ix^{r + 1} = x_ i$ for $i < k$ and $i > n - r - 1$. Finally, we take $x = x^{n - k}$ which finishes the proof. $\square$

Proof. Consider a commutative solid diagram

as in Definition 14.31.1. The map $a$ corresponds to $x_0, \ldots , \hat x_ k, \ldots , x_ n \in X_{n - 1}$ satisfying $d_ i x_ j = d_{j - 1} x_ i$ for $i < j$, $i, j \not= k$. The map $b$ corresponds to an element $y \in Y_ n$ such that $d_ iy = f(x_ i)$ for $i \not= k$. Our task is to produce an $x \in X_ n$ such that $d_ ix = x_ i$ for $i \not= k$ and $f(x) = y$.

Since $f$ is termwise surjective we can find $x \in X_ n$ with $f(x) = y$. Replace $y$ by $0 = y - f(x)$ and $x_ i$ by $x_ i - d_ ix$ for $i \not= k$. Then we see that we may assume $y = 0$. In particular $f(x_ i) = 0$. In other words, we can replace $X$ by $\mathop{\mathrm{Ker}}(f) \subset X$ and $Y$ by $0$. In this case the statement become Lemma 14.31.6. $\square$

Proof. Consider a commutative solid diagram

as in Definition 14.30.1. The map $a$ corresponds to $x_0, \ldots , x_ n \in X_{n - 1}$ satisfying $d_ i x_ j = d_{j - 1} x_ i$ for $i < j$. The map $b$ corresponds to an element $y \in Y_ n$ such that $d_ iy = f(x_ i)$. Our task is to produce an $x \in X_ n$ such that $d_ ix = x_ i$ and $f(x) = y$.

Since $f$ is termwise surjective we can find $x \in X_ n$ with $f(x) = y$. Replace $y$ by $0 = y - f(x)$ and $x_ i$ by $x_ i - d_ ix$. Then we see that we may assume $y = 0$. In particular $f(x_ i) = 0$. In other words, we can replace $X$ by $\mathop{\mathrm{Ker}}(f) \subset X$ and $Y$ by $0$. This works, because by Homology, Lemma 12.13.6 the homology of the chain complex associated to $\mathop{\mathrm{Ker}}(f)$ is zero and hence $\mathop{\mathrm{Ker}}(f) \to 0$ induces a quasi-isomorphism on associated chain complexes.

Since $X$ is a Kan complex (Lemma 14.31.6) we can find $x \in X_ n$ with $d_ i x = x_ i$ for $i = 0, \ldots , n - 1$. After replacing $x_ i$ by $x_ i - d_ ix$ for $i = 0, \ldots , n$ we may assume that $x_0 = x_1 = \ldots = x_{n - 1} = 0$. In this case we see that $d_ i x_ n = 0$ for $i = 0, \ldots , n - 1$. Thus $x_ n \in N(X)_{n - 1}$ and lies in the kernel of the differential $N(X)_{n - 1} \to N(X)_{n - 2}$. Here $N(X)$ is the normalized chain complex associated to $X$, see Section 14.23. Since $N(X)$ is quasi-isomorphic to $s(X)$ (Lemma 14.23.9) and thus acyclic we find $x \in N(X_ n)$ whose differential is $x_ n$. This $x$ answers the question posed by the lemma and we are done. $\square$

Proof. In this proof we will write $H_ n(Z) = H_ n(s(Z)) = H_ n(N(Z))$ when $Z$ is a simplicial abelian group, with $s$ and $N$ as in Section 14.23. Let $\mathbf{Z}[X]$ denote the free abelian group on $X$ viewed as a simplicial set and similarly for $\mathbf{Z}[Y]$. Consider the commutative diagram

of simplicial abelian groups. Since taking the free abelian group on a set is a functor, we see that the horizontal arrow is a homotopy equivalence of simplicial abelian groups, see Lemma 14.28.4. By Lemma 14.27.2 we see that $H_ n(g) : H_ n(\mathbf{Z}[X]) \to H_ n(\mathbf{Z}[Y])$ is bijective for all $n \geq 0$.

Let $\xi \in H_ n(Y)$. By definition of $N(Y)$ we can represent $\xi$ by an element $y \in N(Y_ n)$ whose boundary is zero. This means $y \in Y_ n$ with $d^ n_0(y) = \ldots = d^ n_{n - 1}(y) = 0$ because $y \in N(Y_ n)$ and $d^ n_ n(y) = 0$ because the boundary of $y$ is zero. Denote $0_ n \in Y_ n$ the zero element. Then we see that

is an element with $d^ n_0(\tilde y) = \ldots = d^ n_{n - 1}(\tilde y) = 0$ and $d^ n_ n(\tilde y) = 0$. Thus $\tilde y$ is in $N(\mathbf{Z}[Y])_ n$ has boundary $0$, i.e., $\tilde y$ determines a class $\tilde\xi \in H_ n(\mathbf{Z}[Y])$ mapping to $\xi$. Because $H_ n(\mathbf{Z}[X]) \to H_ n(\mathbf{Z}[Y])$ is bijective we can lift $\tilde\xi$ to a class in $H_ n(\mathbf{Z}[X])$. Looking at the commutative diagram above we see that $\xi$ is in the image of $H_ n(X) \to H_ n(Y)$.

Let $\xi \in H_ n(X)$ be an element mapping to zero in $H_ n(Y)$. Exactly as in the previous parapgraph we can represent $\xi$ by an element $x \in N(X_ n)$ whose boundary is zero, i.e., $d^ n_0(x) = \ldots = d^ n_{n - 1}(x) = d^ n_ n(x) = 0$. In particular, we see that $[x] - [0_ n]$ is an element of $N(\mathbf{Z}[X])_ n$ whose boundary is zero, whence defines a lift $\tilde\xi \in H_ n(\mathbf{Z}[x])$ of $\xi$. The fact that $\xi$ maps to zero in $H_ n(Y)$ means there exists a $y \in N(Y_{n + 1})$ whose boundary is $f_ n(x)$. This means $d^{n + 1}_0(y) = \ldots = d^{n + 1}_ n(y) = 0$ and $d^{n + 1}_{n + 1}(y) = f(x)$. However, this means exactly that $z = [y] - [0_{n + 1}]$ is in $N(\mathbf{Z}[y])_{n + 1}$ and

This proves that $\tilde\xi$ maps to zero in $H_ n(\mathbf{Z}[y])$. As $H_ n(\mathbf{Z}[X]) \to H_ n(\mathbf{Z}[Y])$ is bijective we conclude $\tilde\xi = 0$ and hence $\xi = 0$. $\square$