
14.32 A homotopy equivalence

Suppose that $A$, $B$ are sets, and that $f : A \to B$ is a map. Consider the associated map of simplicial sets

$\xymatrix{ \text{cosk}_0(A) \ar@{=}[r] & \Big(\ldots A \times A \times A \ar[d] \ar@<2ex>[r] \ar@<0ex>[r] \ar@<-2ex>[r] & A \times A \ar[d] \ar@<1ex>[r] \ar@<-1ex>[r] \ar@<1ex>[l] \ar@<-1ex>[l] & A \Big) \ar[d] \ar@<0ex>[l] \\ \text{cosk}_0(B) \ar@{=}[r] & \Big( \ldots B \times B \times B \ar@<2ex>[r] \ar@<0ex>[r] \ar@<-2ex>[r] & B \times B \ar@<1ex>[r] \ar@<-1ex>[r] \ar@<1ex>[l] \ar@<-1ex>[l] & B \Big) \ar@<0ex>[l] }$

See Example 14.19.1. The case $n = 0$ of the following lemma says that this map of simplicial sets is a trivial Kan fibration if $f$ is surjective.

Lemma 14.32.1. Let $f : V \to U$ be a morphism of simplicial sets. Let $n \geq 0$ be an integer. Assume

1. The map $f_ i : V_ i \to U_ i$ is a bijection for $i < n$.

2. The map $f_ n : V_ n \to U_ n$ is a surjection.

3. The canonical morphism $U \to \text{cosk}_ n \text{sk}_ n U$ is an isomorphism.

4. The canonical morphism $V \to \text{cosk}_ n \text{sk}_ n V$ is an isomorphism.

Then $f$ is a trivial Kan fibration.

Proof. Consider a solid diagram

$\xymatrix{ \partial \Delta [k] \ar[r] \ar[d] & V \ar[d] \\ \Delta [k] \ar[r] \ar@{-->}[ru] & U }$

as in Definition 14.30.1. Let $x \in U_ k$ be the $k$-simplex corresponding to the lower horizontal arrow. If $k \leq n$ then the dotted arrow is the one corresponding to a lift $y \in V_ k$ of $x$; the diagram will commute as the other nondegenerate simplices of $\Delta [k]$ are in degrees $< k$ where $f$ is an isomorphism. If $k > n$, then by conditions (3) and (4) we have (using adjointness of skeleton and coskeleton functors)

$\mathop{Mor}\nolimits (\Delta [k], U) = \mathop{Mor}\nolimits (\text{sk}_ n\Delta [k], \text{sk}_ nU) = \mathop{Mor}\nolimits (\text{sk}_ n\partial \Delta [k], \text{sk}_ nU) = \mathop{Mor}\nolimits (\partial \Delta [k], U)$

and similarly for $V$ because $\text{sk}_ n\Delta [k] = \text{sk}_ n\partial \Delta [k]$ for $k > n$. Thus we obtain a unique dotted arrow fitting into the diagram in this case also. $\square$

Let $A, B$ be sets. Let $f^0, f^1 : A \to B$ be maps of sets. Consider the induced maps $f^0, f^1 : \text{cosk}_0(A) \to \text{cosk}_0(B)$ abusively denoted by the same symbols. The following lemma for $n = 0$ says that $f^0$ is homotopic to $f^1$. In fact, the homotopy is given by the map $h : \text{cosk}_0(A) \times \Delta [1] \to \text{cosk}_0(A)$ with components

\begin{eqnarray*} h_ m : A \times \ldots \times A \times \mathop{Mor}\nolimits _{\Delta }([m], [1]) & \longrightarrow & A \times \ldots \times A, \\ (a_0, \ldots , a_ m, \alpha ) & \longmapsto & (f^{\alpha (0)}(a_0), \ldots , f^{\alpha (m)}(a_ m)) \end{eqnarray*}

To check that this works, note that for a map $\varphi : [k] \to [m]$ the induced maps are $(a_0, \ldots , a_ m) \mapsto (a_{\varphi (0)}, \ldots , a_{\varphi (k)})$ and $\alpha \mapsto \alpha \circ \varphi$. Thus $h = (h_ m)_{m \geq 0}$ is clearly a map of simplicial sets as desired.

Lemma 14.32.2. Let $f^0, f^1 : V \to U$ be maps of a simplicial sets. Let $n \geq 0$ be an integer. Assume

1. The maps $f^ j_ i : V_ i \to U_ i$, $j = 0, 1$ are equal for $i < n$.

2. The canonical morphism $U \to \text{cosk}_ n \text{sk}_ n U$ is an isomorphism.

3. The canonical morphism $V \to \text{cosk}_ n \text{sk}_ n V$ is an isomorphism.

Then $f^0$ is homotopic to $f^1$.

First proof. Let $W$ be the $n$-truncated simplicial set with $W_ i = U_ i$ for $i < n$ and $W_ n = U_ n / \sim$ where $\sim$ is the equivalence relation generated by $f^0(y) \sim f^1(y)$ for $y \in V_ n$. This makes sense as the morphisms $U(\varphi ) : U_ n \to U_ i$ corresponding to $\varphi : [i] \to [n]$ for $i < n$ factor through the quotient map $U_ n \to W_ n$ because $f^0$ and $f^1$ are morphisms of simplicial sets and equal in degrees $< n$. Next, we upgrade $W$ to a simplicial set by taking $\text{cosk}_ n W$. By Lemma 14.32.1 the morphism $g : U \to W$ is a trivial Kan fibration. Observe that $g \circ f^0 = g \circ f^1$ by construction and denote this morphism $f : V \to W$. Consider the diagram

$\xymatrix{ \partial \Delta [1] \times V \ar[rr]_{f^0, f^1} \ar[d] & & U \ar[d] \\ \Delta [1] \times V \ar[rr]^ f \ar@{-->}[rru] & & W }$

By Lemma 14.30.2 the dotted arrow exists and the proof is done. $\square$

Second proof. We have to construct a morphism of simplicial sets $h : V \times \Delta [1] \to U$ which recovers $f^ i$ on composing with $e_ i$. The case $n = 0$ was dealt with above the lemma. Thus we may assume that $n \geq 1$. The map $\Delta [1] \to \text{cosk}_1 \text{sk}_1 \Delta [1]$ is an isomorphism, see Lemma 14.19.15. Thus we see that $\Delta [1] \to \text{cosk}_ n \text{sk}_ n \Delta [1]$ is an isomorphism as $n \geq 1$, see Lemma 14.19.10. And hence $V \times \Delta [1] \to \text{cosk}_ n \text{sk}_ n (V \times \Delta [1])$ is an isomorphism too, see Lemma 14.19.12. In other words, in order to construct the homotopy it suffices to construct a suitable morphism of $n$-truncated simplicial sets $h : \text{sk}_ n V \times \text{sk}_ n \Delta [1] \to \text{sk}_ n U$.

For $k = 0, \ldots , n - 1$ we define $h_ k$ by the formula $h_ k(v, \alpha ) = f^0(v) = f^1(v)$. The map $h_ n : V_ n \times \mathop{Mor}\nolimits _{\Delta }([k], [1]) \to U_ n$ is defined as follows. Pick $v \in V_ n$ and $\alpha : [n] \to [1]$:

1. If $\mathop{\mathrm{Im}}(\alpha ) = \{ 0\}$, then we set $h_ n(v, \alpha ) = f^0(v)$.

2. If $\mathop{\mathrm{Im}}(\alpha ) = \{ 0, 1\}$, then we set $h_ n(v, \alpha ) = f^0(v)$.

3. If $\mathop{\mathrm{Im}}(\alpha ) = \{ 1\}$, then we set $h_ n(v, \alpha ) = f^1(v)$.

Let $\varphi : [k] \to [l]$ be a morphism of $\Delta _{\leq n}$. We will show that the diagram

$\xymatrix{ V_{l} \times \mathop{Mor}\nolimits ([l], [1]) \ar[r] \ar[d] & U_{l} \ar[d] \\ V_{k} \times \mathop{Mor}\nolimits ([k], [1]) \ar[r] & U_{k} }$

commutes. Pick $v \in V_{l}$ and $\alpha : [l] \to [1]$. The commutativity means that

$h_ k(V(\varphi )(v), \alpha \circ \varphi ) = U(\varphi )(h_ l(v, \alpha )).$

In almost every case this holds because $h_ k(V(\varphi )(v), \alpha \circ \varphi ) = f^0(V(\varphi )(v))$ and $U(\varphi )(h_ l(v, \alpha )) = U(\varphi )(f^0(v))$, combined with the fact that $f^0$ is a morphism of simplicial sets. The only cases where this does not hold is when either (A) $\mathop{\mathrm{Im}}(\alpha ) = \{ 1\}$ and $l = n$ or (B) $\mathop{\mathrm{Im}}(\alpha \circ \varphi ) = \{ 1\}$ and $k = n$. Observe moreover that necessarily $f^0(v) = f^1(v)$ for any degenerate $n$-simplex of $V$. Thus we can narrow the cases above down even further to the cases (A) $\mathop{\mathrm{Im}}(\alpha ) = \{ 1\}$, $l = n$ and $v$ nondegenerate, and (B) $\mathop{\mathrm{Im}}(\alpha \circ \varphi ) = \{ 1\}$, $k = n$ and $V(\varphi )(v)$ nondegenerate.

In case (A), we see that also $\mathop{\mathrm{Im}}(\alpha \circ \varphi ) = \{ 1\}$. Hence we see that not only $h_ l(v, \alpha ) = f^1(v)$ but also $h_ k(V(\varphi )(v), \alpha \circ \varphi ) = f^1(V(\varphi )(v))$. Thus we see that the relation holds because $f^1$ is a morphism of simplicial sets.

In case (B) we conclude that $l = k = n$ and $\varphi$ is bijective, since otherwise $V(\varphi )(v)$ is degenerate. Thus $\varphi = \text{id}_{[n]}$, which is a trivial case. $\square$

Lemma 14.32.3. Let $A$, $B$ be sets, and that $f : A \to B$ is a map. Consider the simplicial set $U$ with $n$-simplices

$A \times _ B A \times _ B \ldots \times _ B A\ (n + 1 \text{ factors)}.$

see Example 14.3.5. If $f$ is surjective, the morphism $U \to B$ where $B$ indicates the constant simplicial set with value $B$ is a trivial Kan fibration.

Proof. Observe that $U$ fits into a cartesian square

$\xymatrix{ U \ar[d] \ar[r] & \text{cosk}_0(B) \ar[d] \\ B \ar[r] & \text{cosk}_0(A) }$

Since the right vertical arrow is a trivial Kan fibration by Lemma 14.32.1, so is the left by Lemma 14.30.3. $\square$

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