# Kerodon

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### 3.2.5 The Long Exact Sequence of a Fibration

If $(X,x)$ is a pointed Kan complex, then we regard each $\pi _{n}(X,x)$ as a pointed set, with base point given by the homotopy class of the constant map $\Delta ^{n} \rightarrow \{ x\} \subseteq X$ (if $n \geq 1$, then this is the identity element with respect to the group structure on $\pi _{n}(X,x)$). Recall that a diagram of pointed sets

$\cdots \rightarrow ( G_{n+1}, e_{n+1}) \xrightarrow {f_ n} ( G_ n, e_ n ) \xrightarrow { f_{n-1} } (G_{n-1}, e_{n-1} ) \rightarrow \cdots$

is said to be exact if the image of each $f_{n}$ is equal to the fiber $f_{n-1}^{-1} \{ e_{n-1} \} = \{ g \in G_ n: f_{n-1}(g) = e_{n-1} \}$. Our goal in this section is to prove the following:

Theorem 3.2.5.1. Let $f: (X,x) \rightarrow (S,s)$ be a Kan fibration between pointed Kan complexes. Then the sequence of pointed sets

$\cdots \rightarrow \pi _{2}(S,s) \xrightarrow {\partial } \pi _{1}(X_ s, x) \rightarrow \pi _{1}( X, x) \rightarrow \pi _1(S,s) \xrightarrow {\partial } \pi _{0}(X_ s, x) \rightarrow \pi _0( X,x) \rightarrow \pi _0(S,s)$

is exact; here $\partial : \pi _{n+1}(S,s) \rightarrow \pi _{n}(X_ s, x)$ denotes the connecting homomorphism of Construction 3.2.4.3.

Theorem 3.2.5.1 really amounts to three separate assertions, which we will formulate and prove individually (Propositions 3.2.5.2, 3.2.5.4, and 3.2.5.6).

Proposition 3.2.5.2. Let $f: (X,x) \rightarrow (S,s)$ be a Kan fibration between pointed Kan complexes and let $n \geq 0$ be an integer. Then the sequence of pointed sets

$\pi _{n}(X_ s, x) \rightarrow \pi _{n}(X,x) \rightarrow \pi _{n}(S,s)$

is exact.

In the special case $n = 0$, the content of Proposition 3.2.5.2 can be formulated without reference to the base point $x \in X$:

Corollary 3.2.5.3. Let $f: X \rightarrow S$ be a Kan fibration between Kan complexes, let $s$ be a vertex of $S$, and set $X_{s} = \{ s\} \times _{S} X$. Then the image of the map $\pi _0(X_{s} ) \rightarrow \pi _0(X)$ is equal to the fiber of the map $\pi _0(f): \pi _0(X) \rightarrow \pi _0(S)$ over the connected component $[s] \in \pi _0(S)$ determined by the vertex $s$. In other words, a vertex $x \in X$ satisfies $[ f(x) ] = [s]$ in $\pi _0(S)$ if and only if the connected component of $x$ has nonempty intersection with the fiber $X_{s}$.

Proof of Proposition 3.2.5.2. Fix an $n$-simplex $\sigma : \Delta ^{n} \rightarrow X$ such that $\sigma |_{ \operatorname{\partial \Delta }^{n} }$ is the constant map carrying $\operatorname{\partial \Delta }^{n}$ to the base point $x \in X$. We wish to show that the homotopy class $[\sigma ]$ belongs to the image of the map $\pi _{n}(X_ s, x) \rightarrow \pi _{n}(X,x)$ if and only if the image $[f(\sigma )]$ is equal to the base point of $\pi _{n}(S,s)$. The “only if” direction is clear, since the composite map $X_{s} \hookrightarrow X \xrightarrow {f} S$ is equal to the constant map taking the value $s$. For the converse, suppose that $[ f(\sigma ) ]$ is the base point of $\pi _{n}(S,s)$. Then there exists a homotopy $h: \Delta ^{1} \times \Delta ^{n} \rightarrow S$ from $f(\sigma )$ to the constant map $\sigma '_0: \Delta ^{n} \rightarrow \{ s\} \subseteq S$, which is constant when restricted to the boundary $\operatorname{\partial \Delta }^ n$. Since $f$ is a Kan fibration, we can lift $h$ to a homotopy $\widetilde{h}: \Delta ^{1} \times \Delta ^{n} \rightarrow X$ from $\sigma$ to another $n$-simplex $\sigma ': \Delta ^{n} \rightarrow X$, where $\widetilde{h}$ is constant along the boundary $\operatorname{\partial \Delta }^{n}$ and $f( \sigma ') = \sigma '_0$ (Remark 3.1.5.3). Then $\sigma '$ represents a homotopy class $[\sigma '] \in \pi _{n}(X_ s, x)$, and the homotopy $\widetilde{h}$ witnesses that $[\sigma ]$ is equal to the image of $[\sigma ']$ in $\pi _{n}(X,x)$. $\square$

Proposition 3.2.5.4. Let $f: (X,x) \rightarrow (S,s)$ be a Kan fibration between pointed Kan complexes and let $n \geq 0$ be an integer. Then the sequence of pointed sets $\pi _{n+1}(S,s) \xrightarrow { \partial } \pi _{n}(X_ s, x) \rightarrow \pi _{n}(X,x)$ is exact, where $\partial$ is the connecting homomorphism of Construction 3.2.4.3.

In the special case $n=0$, Proposition 3.2.5.4 can also be formulated without reference to the base point $x \in X$.

Corollary 3.2.5.5. Let $f: X \rightarrow S$ be a Kan fibration between Kan complexes, let $s$ be a vertex of $S$, and set $X_{s} = \{ s\} \times _{S} X$. Then two elements of $\pi _0( X_ s )$ have the same image in $\pi _0( X )$ if and only if they belong to the same orbit of the action of the fundamental group $\pi _{1}(S,s)$ (see Variant 3.2.4.5). In other words, the inclusion of Kan complexes $X_{s} \hookrightarrow X$ induces a monomorphism of sets $(\pi _{1}(S,s) \backslash \pi _{0}(X_ s)) \hookrightarrow \pi _0(X)$.

Proof of Proposition 3.2.5.4. Fix an $n$-simplex $\sigma : \Delta ^{n} \rightarrow X_ s$ such that $\sigma |_{ \operatorname{\partial \Delta }^{n} }$ is the constant map carrying $\operatorname{\partial \Delta }^{n}$ to the base point $x \in X_ s$. By construction, the homotopy class $[ \sigma ] \in \pi _{n}(X_ s, x)$ belongs to the image of the connecting homomorphism $\partial : \pi _{n+1}(S,s) \rightarrow \pi _{n}(X_ s, x)$ if and only if there exists an $(n+1)$-simplex $\tau : \Delta ^{n+1} \rightarrow S$ such that $\tau |_{ \operatorname{\partial \Delta }^{n+1} }$ is the constant map taking the value $s$ and $\sigma$ is incident to $\tau$, in the sense of Definition 3.2.4.1. This condition is equivalent to the existence of an $(n+1)$-simplex $\widetilde{\tau }: \Delta ^{n+1} \rightarrow X$ satisfying $d_0( \widetilde{\tau } ) = \sigma$ and $d_ i( \widetilde{\tau } )$ is equal to the constant map $e: \Delta ^{n} \rightarrow \{ x\} \subseteq X$ for $1 \leq i \leq n+1$. In other words, it is equivalent to the assertion that the tuple of $n$-simplices of $X$ $( \sigma , e, e, \ldots , e)$ bounds, in the sense of Notation 3.2.3.1. For $n \geq 1$, this is equivalent to the vanishing of the image of $[\sigma ]$ in the homotopy group $\pi _{n}(X,x)$ (Theorem 3.2.2.10). When $n=0$, it is equivalent to the equality $[\sigma ] = [x]$ in $\pi _0(X)$ by virtue of Remark 1.3.6.13. $\square$

Proposition 3.2.5.6. Let $f: (X,x) \rightarrow (S,s)$ be a Kan fibration between pointed Kan complexes and let $n \geq 0$ be an integer. Then the sequence of pointed sets $\pi _{n+1}(X,x) \xrightarrow { \pi _{n+1}(f)} \pi _{n+1}(S,s) \xrightarrow {\partial } \pi _{n}(X_ s,x)$ is exact, where $\partial$ is the connecting homomorphism of Construction 3.2.4.3.

Corollary 3.2.5.7. Let $f: (X,x) \rightarrow (S,s)$ be a Kan fibration between pointed Kan complexes. Then the image of the induced map $\pi _{1}(f): \pi _1(X,x) \rightarrow \pi _{1}(S,s)$ is equal to the stabilizer of $[x] \in \pi _0(X_ s)$ (with respect to the action of $\pi _{1}(S,s)$ on $\pi _0(X_ s)$ supplied by Variant 3.2.4.5.

Proof of Proposition 3.2.5.6. Fix an $(n+1)$-simplex $\tau : \Delta ^{n+1} \rightarrow S$ for which $\tau |_{ \operatorname{\partial \Delta }^{n+1} }$ is the constant map taking the value $s$. By construction, the connecting homomorphism $\partial : \pi _{n+1}(S,s) \rightarrow \pi _{n}(X_ s, x)$ carries $[ \tau ]$ to the base point of $\pi _{n}(X_ s,x)$ if and only if the constant map $e: \Delta ^{n} \rightarrow \{ x\} \hookrightarrow X_{s}$ is incident to $\tau$, in the sense of Definition 3.2.4.1. This is equivalent to the requirement that $\tau$ can be lifted to a map $\widetilde{\tau }: \Delta ^{n+1} \rightarrow X$ for which $\widetilde{\tau }|_{ \operatorname{\partial \Delta }^{n+1} }$ is the constant map taking the value $x$, which clearly implies that that $[\tau ]$ belongs to the image of the map $\pi _{n+1}(f): \pi _{n+1}(X,x) \rightarrow \pi _{n+1}(S,s)$. To prove the reverse implication, suppose that $[\tau ]$ belongs to the image of $\pi _{n+1}(f)$, so that we can write $[ \tau ] = [ f( \widetilde{\tau }' ) ]$ for some map $\widetilde{\tau }': \Delta ^{n+1} \rightarrow X$ for which $\widetilde{\tau }'|_{ \operatorname{\partial \Delta }^{n+1} }$ is the constant map taking the value $x$. It follows that there is a homotopy $h: \Delta ^{1} \times \Delta ^{n+1} \rightarrow S$ from $f( \widetilde{\tau }' )$ to $\tau$ which is constant along the boundary $\operatorname{\partial \Delta }^{n+1}$. Since $f$ is a Kan fibration, we can lift $h$ to a map $\widetilde{h}: \Delta ^{1} \times \Delta ^{n+1} \rightarrow X$ such that $h|_{ \{ 0\} \times \Delta ^{n+1} } = \widetilde{\tau }'$ and $h|_{ \Delta ^{1} \times \operatorname{\partial \Delta }^{n+1} }$ is the constant map taking the value $x$ (Remark 3.1.5.3). The restriction $\widetilde{\tau } = h|_{ \{ 1\} \times \Delta ^{n+1} }$ then satisfies $f( \widetilde{\tau } ) = \tau$ and $\widetilde{\tau }|_{ \operatorname{\partial \Delta }^{n+1} }$ is the constant map taking the value $x$. $\square$