14.31 Kan fibrations

Let $n$, $k$ be integers with $0 \leq k \leq n$ and $1 \leq n$. Let $\sigma _0, \ldots , \sigma _ n$ be the $n + 1$ faces of the unique nondegenerate $n$-simplex $\sigma$ of $\Delta [n]$, i.e., $\sigma _ i = d_ i\sigma$. We let

$\Lambda _ k[n] \subset \Delta [n]$

be the $k$th horn of the $n$-simplex $\Delta [n]$. It is the simplicial subset of $\Delta [n]$ generated by $\sigma _0, \ldots , \hat\sigma _ k, \ldots , \sigma _ n$. In other words, the image of the displayed inclusion contains all the nondegenerate simplices of $\Delta [n]$ except for $\sigma$ and $\sigma _ k$.

Definition 14.31.1. A map $X \to Y$ of simplicial sets is called a Kan fibration if for all $k, n$ with $1 \leq n$, $0 \leq k \leq n$ and any commutative solid diagram

$\xymatrix{ \Lambda _ k[n] \ar[r] \ar[d] & X \ar[d] \\ \Delta [n] \ar[r] \ar@{-->}[ru] & Y }$

a dotted arrow exists making the diagram commute. A Kan complex is a simplicial set $X$ such that $X \to *$ is a Kan fibration, where $*$ is the constant simplicial set on a singleton.

Note that $\Lambda _ k[n]$ is always nonempty. Thus a morphism from the empty simplicial set to any simplicial set is always a Kan fibration. It follows from Lemma 14.30.2 that a trivial Kan fibration is a Kan fibration.

Lemma 14.31.2. Let $f : X \to Y$ be a Kan fibration of simplicial sets. Let $Y' \to Y$ be a morphism of simplicial sets. Then $X \times _ Y Y' \to Y'$ is a Kan fibration.

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

Lemma 14.31.3. The composition of two Kan fibrations is a Kan fibration.

Proof. Omitted. $\square$

Lemma 14.31.4. Let $\ldots \to U^2 \to U^1 \to U^0$ be a sequence of Kan fibrations. Let $U = \mathop{\mathrm{lim}}\nolimits U^ t$ defined by taking $U_ n = \mathop{\mathrm{lim}}\nolimits U_ n^ t$. Then $U \to U^0$ is a Kan fibration.

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

Lemma 14.31.5. Let $X_ i \to Y_ i$ be a set of Kan fibrations. Then $\prod X_ i \to \prod Y_ i$ is a Kan fibration.

Proof. Omitted. $\square$

The following lemma is due to J.C. Moore, see .

Lemma 14.31.6. Let $X$ be a simplicial group. Then $X$ is a Kan complex.

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

$y^ r = s_{r + 1}((d_{r + 1}u^ r)^{-1}x_{r + 1}),\quad u^{r + 1} = u^ r y^ r.$

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

$d_ i x^ r = x_ i\quad \text{for }i < k\text{ and }i > n - r.$

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

$z^ r = s_{n - r - 1}((d_{n - r}x^ r)^{-1}x_{n - r}),\quad x^{r + 1} = x^ rz^ r$

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. $\square$

Lemma 14.31.7. Let $f : X \to Y$ be a homomorphism of simplicial abelian groups which is termwise surjective. Then $f$ is a Kan fibration of simplicial sets.

Proof. Consider a commutative solid diagram

$\xymatrix{ \Lambda _ k[n] \ar[r]_ a \ar[d] & X \ar[d] \\ \Delta [n] \ar[r]^ b \ar@{-->}[ru] & Y }$

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. $\square$

Lemma 14.31.8. Let $f : X \to Y$ be a homomorphism of simplicial abelian groups which is termwise surjective and induces a quasi-isomorphism on associated chain complexes. Then $f$ is a trivial Kan fibration of simplicial sets.

Proof. Consider a commutative solid diagram

$\xymatrix{ \partial \Delta [n] \ar[r]_ a \ar[d] & X \ar[d] \\ \Delta [n] \ar[r]^ b \ar@{-->}[ru] & Y }$

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 12.12.6 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 14.23.8) 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. $\square$

Lemma 14.31.9. Let $f : X \to Y$ be a map of simplicial abelian groups. If $f$ is termwise surjective1 and a homotopy equivalence of simplicial sets, then $f$ induces a quasi-isomorphism of associated chain complexes.

Proof. By assumption there exists a map $g : Y \to X$ of simplicial sets, a homotopy $h : X \times \Delta [1] \to X$ between $g \circ f$ and $\text{id}_ X$, and a homotopy $h' : Y \times \Delta [1] \to Y$ between $f \circ g$ and $\text{id}_ Y$. During this proof we will write $H_ n(X) = H_ n(s(X)) = H_ n(N(X))$, see Section 14.23.

Note that $H_0(X)$ is the cokernel of the difference map $d_1 - d_0 : X_1 \to X_0$. Observe that $x \in X_0$ corresponds to a morphism $\Delta [0] \to X$. Composing $h$ with the induced map $\Delta [0] \times \Delta [1] \to X \times \Delta [1]$ we see that $x$ and $g(f(x))$ are equal to $d_0x'$ and $d_1x'$ for some $x' \in X_1$. Similarly for $y \in Y_0$. We conclude that $f$ defines a bijection $H_0(X) \to H_0(Y)$.

Let $n \geq 1$. Consider the simplicial set $S$ which is the pushout of

$\xymatrix{ \partial \Delta [n] \ar[r] \ar[d] & {*} \ar[d] \\ \Delta [n] \ar[r] & S }$

Concretely, we take

$S_ k = \{ \varphi : [k] \to [n] \mid \varphi \text{ is surjective}\} \amalg \{ *\} .$

Denote $E = \mathbf{Z}[S]$ the free abelian group on $S$. The inclusion $\Delta [0] \to S$ coming from $* \in S_0$ determines an injection $K(\mathbf{Z}, 0) \to E$ whose cokernel is the object $K(\mathbf{Z}, n)$, i.e., we have a short exact sequence

$0 \to K(\mathbf{Z}, 0) \to E \to K(\mathbf{Z}, n) \to 0$

See Definition 14.22.3 and the description of the Eilenberg-Maclane objects in Lemma 14.22.2. Note that the extension above is split, for example because the element $\xi = [\text{id}_{[n]}] - [*] \in E_ n$ satisfies $d_ i\xi = 0$ and maps to the “generator” of $K(\mathbf{Z}, n)$. We have

$\mathop{Mor}\nolimits _{\text{Simp}(\textit{Sets})}(S, X) = \mathop{Mor}\nolimits _{\text{Simp}(\textit{Ab})}(E, X) = X_0 \times \bigcap \nolimits _{i = 0, \ldots , n} \mathop{\mathrm{Ker}}(d_ i : X_ n \to X_{n - 1})$

This uses the choice of our splitting above and the description of morphisms out of Eilenberg-Maclane objects given in Lemma 14.22.2. Note that we can think of $\bigcap \nolimits _{i = 0, \ldots , n} \mathop{\mathrm{Ker}}(d_ i : X_ n \to X_{n - 1})$ as the cycles in degree $n$ in the normalized chain complex associated to $X$, see Section 14.23. If two maps $a, b : S \to X$ are homotopic (as maps of simplicial sets), then the corresponding maps $a', b' : E \to X$ are homotopic as maps of simplicial abelian groups (because taking the free abelian group on is a functor). Thus if $a$, resp. $b$ correspond to $(a_0, a_ n)$, resp. $(b_0, b_ n)$ in the formula above, then $a_0$ and $b_0$ define the same element of $H_0(X)$ and $a_ n$ and $b_ n$ define the same class in $H_ n(X)$. See Lemma 14.27.1.

We come the final arguments of the proof. An element $y$ of $H_ n(Y)$ can be represented by an element $y_ n$ in $\bigcap \nolimits _{i = 0, \ldots , n} \mathop{\mathrm{Ker}}(d_ i : Y_ n \to Y_{n - 1})$. Let $a : S \to Y$ be the map of simplicial sets corresponding to $(0, y_ n)$. Then $b = g \circ a$ corresponds to some $(b_0, b_ n)$ as above for $X$. Using the homotopy $h'$ we see $(f(b_0), f(b_ n))$ and $(0, y_ n)$ come from homotopic maps $S \to Y$ and hence $y_ n$ and $f(b_ n)$ define the same element of $H_ n(Y)$. Clearly this shows that $H_ n(f)$ is surjective. Conversely, suppose $x_ n$ in $\bigcap \nolimits _{i = 0, \ldots , n} \mathop{\mathrm{Ker}}(d_ i : X_ n \to X_{n - 1})$ and $f(x_ n) = d(y')$ with $y' \in N(Y_{n + 1})$. Since $f$ is termwise surjective so is the induced map $f : N(X_{n + 1}) \to N(Y_{n + 1})$ (see Lemma 14.23.6). Thus we can pick $x' \in N(X_{n + 1})$ mapping to $y'$. After replacing $x_ n$ by $x_ n - d(x')$ we reach the point where $f(x_ n) = 0$. This means that the morphism $a : S \to X$ corresponding to $(0, x_ n)$ has the property that $f \circ a$ is the constant morphism with value $0$ in $Y$. Hence $g \circ f \circ a$ is also a constant morphism, i.e., corresponds to a pair $(b_0, 0)$. Since as before $x_ n$ and $0$ represent the same element of $H_ n(X)$ we conclude. $\square$

[1] This assumption is not necessary. Also the proof as currently given is not the right one. A better proof is to define the homotopy groups of Kan complex and show that these are equal to the homology groups of the associated complex for a simplicial abelian group.

Comment #3997 by Ajneet Dhillon on

There is a very minor typo, "This a morphism from an empty..." should read "Thus a morphism..."

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