8.4 Fiber sequences and the long exact sequence

If the codomain of a function $f:X\to Y$ is equipped with a basepoint $y_0:Y$, then we refer to the fiber $F:\equiv {\mathrm{𝖿𝗂𝖻}}_f(y_0)$ of $f$ over $y_0$ as the fiber of $f$. (If $Y$ is connected, then $F$ is determined up to mere equivalence; see ex:unique-fiber.) We now show that if $X$ is also pointed and $f$ preserves basepoints, then there is a relation between the homotopy groups of $F$, $X$, and $Y$ in the form of a long exact sequence. We derive this by way of the fiber sequence associated to such an $f$.

Definition 8.4.1.

A pointed map between pointed types $\mathrm{(}X\mathrm{,}{x}_{\mathrm{0}}\mathrm{)}$ and $\mathrm{(}Y\mathrm{,}{y}_{\mathrm{0}}\mathrm{)}$ is a map $f\mathrm{:}X\mathrm{\to }Y$ together with a path $f_{\mathrm{0}}\mathrm{:}f\mathit{}\mathrm{(}{x}_{\mathrm{0}}\mathrm{)}\mathrm{=}{y}_{\mathrm{0}}$.

For any pointed types $(X,x_0)$ and $(Y,y_0)$, there is a pointed map $(\lambda x.y_0):X\to Y$ which is constant at the basepoint. We call this the zero map and sometimes write it as $0:X\to Y$.

Recall that every pointed type $(X,x_0)$ has a loop space $\mathrm{\Omega }(X,x_0)$. We now note that this operation is functorial on pointed maps.

Definition 8.4.2.

Given a pointed map between pointed types $f\mathrm{:}X\mathrm{\to }Y$, we define a pointed map $\mathrm{\Omega }\mathit{}f\mathrm{:}\mathrm{\Omega }\mathit{}X\mathrm{\to }\mathrm{\Omega }\mathit{}Y$ by

$$(\mathrm{\Omega }f)(p):\equiv f_0{}^{-1}f\left(p\right)f_0.$$

The path ${\mathrm{(}\mathrm{\Omega }\mathit{}f\mathrm{)}}_{\mathrm{0}}\mathrm{:}\mathrm{(}\mathrm{\Omega }\mathit{}f\mathrm{)}\mathit{}\mathrm{(}{\mathrm{refl}}_{x_{\mathrm{0}}}\mathrm{)}\mathrm{=}{\mathrm{refl}}_{y_{\mathrm{0}}}$, which exhibits $\mathrm{\Omega }\mathit{}f$ as a pointed map, is the obvious path of type

$$f_0{}^{-1}f\left({\mathrm{𝗋𝖾𝖿𝗅}}_{x_0}\right)f_0={\mathrm{𝗋𝖾𝖿𝗅}}_{y_0}.$$

There is another functor on pointed maps, which takes $f:X\to Y$ to ${\mathrm{𝗉𝗋}}_1:{\mathrm{𝖿𝗂𝖻}}_f(y_0)\to X$. When $f$ is pointed, we always consider ${\mathrm{𝖿𝗂𝖻}}_f(y_0)$ to be pointed with basepoint $(x_0,f_0)$, in which case ${\mathrm{𝗉𝗋}}_1$ is also a pointed map, with witness ${({\mathrm{𝗉𝗋}}_1)}_0:\equiv {\mathrm{𝗋𝖾𝖿𝗅}}_{x_0}$. Thus, this operation can be iterated.

Definition 8.4.3.

The fiber sequence of a pointed map $f\mathrm{:}X\mathrm{\to }Y$ is the infinite sequence of pointed types and pointed maps

defined recursively by

$$X^{(0)}:\equiv Y\mathit{\hspace{1em}\hspace{1em}}{X}^{(1)}:\equiv X\mathit{\hspace{1em}\hspace{1em}}{f}^{(0)}:\equiv f\mathit{\hspace{1em}\hspace{1em}}$$

and

$X^{(n+1)}$ $:\equiv {\mathrm{𝖿𝗂𝖻}}_{f^{(n-1)}}(x_0^{(n-1)})$

$f^{(n)}$ $:\equiv {\mathrm{𝗉𝗋}}_1$ $:X^{(n+1)}\to X^{(n)}.$

where $x_{\mathrm{0}}^{\mathrm{(}n\mathrm{)}}$ denotes the basepoint of $X^{\mathrm{(}n\mathrm{)}}$, chosen recursively as above.

Thus, any adjacent pair of maps in this fiber sequence is of the form

In particular, we have $f^{(n-1)}\circ f^{(n)}=0$. We now observe that the types occurring in this sequence are the iterated loop spaces of the base space $Y$, the total space $X$, and the fiber $F:\equiv {\mathrm{𝖿𝗂𝖻}}_f(y_0)$, and similarly for the maps.

Lemma 8.4.4.

Let $f\mathrm{:}X\mathrm{\to }Y$ be a pointed map of pointed spaces. Then:

1. The fiber of $f^{(1)}:\equiv {\mathrm{𝗉𝗋}}_1:{\mathrm{𝖿𝗂𝖻}}_f(y_0)\to X$ is equivalent to $\mathrm{\Omega }Y$.

2. Similarly, the fiber of $f^{(2)}:\mathrm{\Omega }Y\to {\mathrm{𝖿𝗂𝖻}}_f(y_0)$ is equivalent to $\mathrm{\Omega }X$.

3. Under these equivalences, the map $f^{(3)}:\mathrm{\Omega }X\to \mathrm{\Omega }Y$ is identified with $\mathrm{\Omega }f\circ {(\mathit{\text{–}})}^{-1}$.

Proof.

For 1, we have

${\mathrm{𝖿𝗂𝖻}}_{f^{(1)}}(x_0)$ $:\equiv {\displaystyle \sum _{z:{\mathrm{𝖿𝗂𝖻}}_f(y_0)}}({\mathrm{𝗉𝗋}}_1(z)=x_0)$

$\simeq {\displaystyle \sum _{(x:A)}}{\displaystyle \sum _{(p:f(x)=y_0)}}(x=x_0)$ (by ex:sigma-assoc)

$\simeq (f(x_0)=y_0)$ (as ${\sum }_{(x:A)}(x=x_0)$ is contractible)

$\simeq (y_0=y_0)$ (by $(f_0\text{–})$)

$\equiv \mathrm{\Omega }Y.$

Tracing through, we see that this equivalence sends $((x,p),q)$ to $f_0{}^{-1}f\left(q^{-1}\right)p$, while its inverse sends $r:y_0=y_0$ to $((x_0,f_{0}r),{\mathrm{𝗋𝖾𝖿𝗅}}_{x_0})$. In particular, the basepoint $((x_0,f_0),{\mathrm{𝗋𝖾𝖿𝗅}}_{x_0})$ of ${\mathrm{𝖿𝗂𝖻}}_{f^{(1)}}(x_0)$ is sent to $f_0{}^{-1}f\left(\mathrm{𝗋𝖾𝖿𝗅}_{x_0}{}^{-1}\right)f_0$, which equals ${\mathrm{𝗋𝖾𝖿𝗅}}_{y_0}$. Hence this equivalence is a pointed map (see ex:pointed-equivalences). Moreover, under this equivalence, $f^{(2)}$ is identified with $\lambda r.(x_0,f_{0}r):\mathrm{\Omega }Y\to {\mathrm{𝖿𝗂𝖻}}_f(y_0)$.

item:fibseq2 follows immediately by applying 1 to $f^{(1)}$ in place of $f$. Since ${(f^{(1)})}_0:\equiv {\mathrm{𝗋𝖾𝖿𝗅}}_{x_0}$, under this equivalence $f^{(3)}$ is identified with the map $\mathrm{\Omega }X\to {\mathrm{𝖿𝗂𝖻}}_{f^{(1)}}(x_0)$ defined by $s\mapsto ((x_0,f_0),s)$. Thus, when we compose with the previous equivalence ${\mathrm{𝖿𝗂𝖻}}_{f^{(1)}}(x_0)\simeq \mathrm{\Omega }Y$, we see that $s$ maps to $f_0{}^{-1}f\left(s^{-1}\right)f_0$, which is by definition $(\mathrm{\Omega }f)(s^{-1})$, giving 3. ∎

Thus, the fiber sequence of $f:X\to Y$ can be pictured as:

where the minus signs denote composition with path inversion ${(\text{–})}^{-1}$. Note that by ex:ap-path-inversion, we have

$$\mathrm{\Omega }\left(\mathrm{\Omega }f\circ {(\text{–})}^{-1}\right)\circ {(\text{–})}^{-1}={\mathrm{\Omega }}^{2}f\circ {(\text{–})}^{-1}\circ {(\text{–})}^{-1}={\mathrm{\Omega }}^{2}f.$$

Thus, there are minus signs on the $k$-fold loop maps whenever $k$ is odd.

From this fiber sequence we will deduce an exact sequence of pointed sets. Let $A$ and $B$ be sets and $f:A\to B$ a function, and recall from defn:modal-image the definition of the image $\mathrm{𝗂𝗆}(f)$, which can be regarded as a subset of $B$:

$$\mathrm{𝗂𝗆}(f):\equiv b:B|\exists (a:A).f(a)=b.$$

If $A$ and $B$ are moreover pointed with basepoints $a_0$ and $b_0$, and $f$ is a pointed map, we define the kernel of $f$ to be the following subset of $A$:

$$\ker (f):\equiv x:A|f(x)=b_0.$$

Of course, this is just the fiber of $f$ over the basepoint $b_0$; it a subset of $A$ because $B$ is a set.

Note that any group is a pointed set, with its unit element as basepoint, and any group homomorphism is a pointed map. In this case, the kernel and image agree with the usual notions from group theory.

Definition 8.4.5.

An exact sequence of pointed sets is a (possibly bounded) sequence of pointed sets and pointed maps:

such that for every $n$, the image of $f^{\mathrm{(}n\mathrm{)}}$ is equal, as a subset of $A^{\mathrm{(}n\mathrm{)}}$, to the kernel of $f^{\mathrm{(}n\mathrm{-}\mathrm{1}\mathrm{)}}$. In other words, for all $a\mathrm{:}{A}^{\mathrm{(}n\mathrm{)}}$ we have

$$(f^{(n-1)}(a)=a_0^{(n-1)})\iff \exists (b:A^{(n+1)}).(f^{(n)}(b)=a).$$

where $a_{\mathrm{0}}^{\mathrm{(}n\mathrm{)}}$ denotes the basepoint of $A^{\mathrm{(}n\mathrm{)}}$.

Usually, most or all of the pointed sets in an exact sequence are groups, and often abelian groups. When we speak of an exact sequence of groups, it is assumed moreover that the maps are group homomorphisms and not just pointed maps.

Theorem 8.4.6.

Let $f\mathrm{:}X\mathrm{\to }Y$ be a pointed map between pointed spaces with fiber $F\mathrm{:}\mathrm{\equiv }{\mathrm{fib}}_f\mathrm{(}{y}_{\mathrm{0}}\mathrm{)}$. Then we have the following long exact sequence, which consists of groups except for the last three terms, and abelian groups except for the last six.

Proof.

We begin by showing that the 0-truncation of a fiber sequence is an exact sequence of pointed sets. Thus, we need to show that for any adjacent pair of maps in a fiber sequence:

with $g:\equiv {\mathrm{𝗉𝗋}}_1$, the sequence

is exact, i.e. that $\mathrm{𝗂𝗆}(\parallel g{\parallel }_0)\subseteq \ker (\parallel f{\parallel }_0)$ and $\ker (\parallel f{\parallel }_0)\subseteq \mathrm{𝗂𝗆}(\parallel g{\parallel }_0)$.

The first inclusion is equivalent to $\parallel g{\parallel }_0\circ \parallel f{\parallel }_0=0$, which holds by functoriality of ${\parallel \text{–}\parallel }_0$ and the fact that $g\circ f=0$. For the second, we assume $w^{\prime }:\parallel W{\parallel }_0$ and $p^{\prime }:\parallel f{\parallel }_0(w^{\prime })=|z_0{|}_0$ and show there merely exists $t:{\mathrm{𝖿𝗂𝖻}}_f(z_0)$ such that $g(t)=w^{\prime }$. Since our goal is a mere proposition, we can assume that $w^{\prime }$ is of the form ${\left|w\right|}_0$ for some $w:W$. Now by thm:path-truncation, $p^{\prime }:|f(w){|}_0=|z_0{|}_0$ yields $p^{\prime \prime }:\parallel f(w)=z_0{\parallel }_{-1}$, so by a further truncation induction we may assume some $p:f(w)=z_0$. But now we have $|(w,p){|}_0:|{\mathrm{𝖿𝗂𝖻}}_f(z_0){|}_0$ whose image under ${\parallel g\parallel }_0$ is $|w{|}_0\equiv w^{\prime }$, as desired.

Thus, applying ${\parallel \text{–}\parallel }_0$ to the fiber sequence of $f$, we obtain a long exact sequence involving the pointed sets ${\pi }_k(F)$, ${\pi }_k(X)$, and ${\pi }_k(Y)$ in the desired order. And of course, ${\pi }_k$ is a group for $k\ge 1$, being the 0-truncation of a loop space, and an abelian group for $k\ge 2$ by the Eckmann–Hilton argument (thm:EckmannHilton). Moreover, thm:fiber-of-the-fiber allows us to identify the maps ${\pi }_k(F)\to {\pi }_k(X)$ and ${\pi }_k(X)\to {\pi }_k(Y)$ in this exact sequence as ${(-1)}^k{\pi }_k(i)$ and ${(-1)}^k{\pi }_k(f)$ respectively.

More generally, every map in this long exact sequence except the last three is of the form ${\parallel \mathrm{\Omega }h\parallel }_0$ or ${\parallel -\mathrm{\Omega }h\parallel }_0$ for some $h$. In the former case it is a group homomorphism, while in the latter case it is a homomorphism if the groups are abelian; otherwise it is an “anti-homomorphism”. However, the kernel and image of a group homomorphism are unchanged when we replace it by its negative, and hence so is the exactness of any sequence involving it. Thus, we can modify our long exact sequence to obtain one involving ${\pi }_k(i)$ and ${\pi }_k(f)$ directly and in which all the maps are group homomorphisms (except the last three). ∎

The usual properties of exact sequences of abelian groups can be proved as usual. In particular we have:

Lemma 8.4.7.

Suppose given an exact sequence of abelian groups:

1. If $K=0$, then $f$ is injective.

2. If $Q=0$, then $f$ is surjective.

3. If $K=Q=0$, then $f$ is an isomorphism.

Proof.

Since the kernel of $f$ is the image of $K\to G$, if $K=0$ then the kernel of $f$ is $\{0\}$; hence $f$ is injective because it’s a group morphism. Similarly, since the image of $f$ is the kernel of $H\to Q$, if $Q=0$ then the image of $f$ is all of $H$, so $f$ is surjective. Finally, 3 follows from 1 and 2 by thm:mono-surj-equiv. ∎

As an immediate application, we can now quantify in what way $n$-connectedness of a map is stronger than inducing an equivalence on $n$-truncations.

Corollary 8.4.8.

Let $f\mathrm{:}A\mathrm{\to }B$ be $n$-connected and $a\mathrm{:}A$, and define $b\mathrm{:}\mathrm{\equiv }f\mathrm{(}a\mathrm{)}$. Then:

1. If $k\le n$, then ${\pi }_k(f):{\pi }_k(A,a)\to {\pi }_k(B,b)$ is an isomorphism.

2. If $k=n+1$, then ${\pi }_k(f):{\pi }_k(A,a)\to {\pi }_k(B,b)$ is surjective.

Proof.

As part of the long exact sequence, for each $k$ we have an exact sequence

Now since $f$ is $n$-connected, ${\parallel {\mathrm{𝖿𝗂𝖻}}_f(b)\parallel }_n$ is contractible. Therefore, if $k\le n$, then ${\pi }_k({\mathrm{𝖿𝗂𝖻}}_f(b))=\parallel {\mathrm{\Omega }}^k({\mathrm{𝖿𝗂𝖻}}_f(b)){\parallel }_0={\mathrm{\Omega }}^k(\parallel {\mathrm{𝖿𝗂𝖻}}_f(b){\parallel }_k)$ is also contractible. Thus, ${\pi }_k(f)$ is an isomorphism for $k\le n$ by thm:ses3, while for $k=n+1$ it is surjective by thm:ses2. ∎

In sec:whitehead we will see that the converse of thm:conn-pik also holds.

Title	8.4 Fiber sequences and the long exact sequence