Fibrations, Homotopy Groups, and Exact Sequences

Proof. Let $U\ni b$ be evenly covered with sheets $V_i$ and let $x\in p^{-1}(b)$. Then $x$ lies in exactly one sheet $V_i$, because the sheets are disjoint and cover $p^{-1}(U)\supseteq p^{-1}(b)$. Since $p{|}_{V_i}$ is injective, $V_i\cap p^{-1}(b)=\{x\}$, so $\{x\}$ is open in the subspace $p^{-1}(b)$. ◻

Proof. Let $A:=\{z\in Z:f(z)=g(z)\}$; it contains $z_0$. We show that $A$ and $Z\setminus A$ are both open, so that connectedness forces $A=Z$.

Let $z\in Z$ and let $U$ be an evenly covered neighborhood of $p(f(z))$, with sheets $V_i$. Let $V$ and $V^{\prime }$ be the sheets containing $f(z)$ and $g(z)$ respectively, and put $W:=f^{-1}(V)\cap g^{-1}(V^{\prime })$, an open neighborhood of $z$.

If $z\in A$ then $V=V^{\prime }$, and for $w\in W$ both $f(w)$ and $g(w)$ lie in $V$ with $p(f(w))=p(g(w))$; since $p{|}_V$ is injective, $f(w)=g(w)$. So $W\subseteq A$.

If $z\notin A$ then $f(z)\ne g(z)$. Either $V\ne V^{\prime }$, and then for $w\in W$ the points $f(w)\in V$ and $g(w)\in V^{\prime }$ lie in disjoint sets, so $f(w)\ne g(w)$; or $V=V^{\prime }$, and then injectivity of $p{|}_V$ together with $pf(z)=pg(z)$ would give $f(z)=g(z)$, which is excluded. So $W\subseteq Z\setminus A$. ◻

Proof. For each $x\in X$ choose $U_x\in \mathcal{U}$ with $x\in U_x$ and with $B(x,2{\epsilon }_x)\subseteq U_x$. The balls $B(x,{\epsilon }_x)$ cover $X$, so by compactness finitely many $B(x_1,{\epsilon }_1),\dots ,B(x_k,{\epsilon }_k)$ do. Put . Let $A\subseteq X$ have diameter less than $\delta$ and pick $a\in A$. Then $a\in B(x_j,{\epsilon }_j)$ for some $j$, and every $a^{\prime }\in A$ has , so $A\subseteq B(x_j,2{\epsilon }_j)\subseteq U_{x_j}$. ◻

Proof. Uniqueness is lemma 188.4 with $Z=[0,1]$, which is connected by lemma 187.10.

For existence, cover $B$ by evenly covered open sets and pull back along $\gamma$: the sets ${\gamma }^{-1}(U)$ form an open cover of the compact metric space $[0,1]$ (lemma 187.13). Let $\delta$ be a Lebesgue number (lemma 188.5) and choose with , so that $\gamma ([t_j,t_{j+1}])\subseteq U_j$ for some evenly covered $U_j$.

Construct $\tilde{\gamma }$ on $[0,t_j]$ by induction on $j$. For $j=0$ set $\tilde{\gamma }(0)=x_0$. Given the lift on $[0,t_j]$, let $V$ be the unique sheet over $U_j$ containing $\tilde{\gamma }(t_j)$, which exists because $p(\tilde{\gamma }(t_j))=\gamma (t_j)\in U_j$, and define on $[t_j,t_{j+1}]$ $$\tilde{\gamma }(s):=(p{|}_V{)}^{-1}(\gamma (s)),$$ which is continuous as a composite and agrees with the previous stage at $t_j$. The two closed pieces $[0,t_j]$ and $[t_j,t_{j+1}]$ satisfy the hypotheses of lemma 187.11, so the extension is continuous, and $p\circ \tilde{\gamma }=\gamma$ holds on the new piece by construction. ◻

Proof. Uniqueness is again lemma 188.4, applied on the connected space $I^n\times [0,1]$; two lifts agreeing on $I^n\times \{0\}$ agree at one point, hence everywhere.

For existence, first fix $z\in I^n$. The proof is written for $n=1$; for general $n$ replace each occurrence of a subinterval of $I$ by a subcube of $I^n$ and each occurrence of $|s-s^{\prime }|$ by the maximum coordinate distance, leaving every other word unchanged.

Cover $I\times I$ by the open sets $H^{-1}(U)$ for $U$ evenly covered, and let $\delta$ be a Lebesgue number for that cover of the compact metric space $I\times I$. Choose and with all gaps below $\delta /2$, so that each closed rectangle $R_{ij}:=[s_i,s_{i+1}]\times [t_j,t_{j+1}]$ has diameter below $\delta$ and hence $H(R_{ij})\subseteq U_{ij}$ for some evenly covered $U_{ij}$.

Order the rectangles lexicographically by $(j,i)$ and lift them in that order, maintaining the invariant that $\tilde{H}$ has been defined and is continuous on $$A_{ij}:=(I\times [0,t_j])\cup ([0,s_i]\times [t_j,t_{j+1}]),$$ that it lifts $H$ there, and that it restricts to $h_0$ on $I\times \{0\}$. At the first rectangle, $A_{00}=I\times \{0\}$ and the invariant holds by hypothesis.

At the step for $R_{ij}$, the set $C:=A_{ij}\cap R_{ij}$ is the union of the bottom edge of $R_{ij}$ and, when , its left edge; in both cases $C$ is connected and nonempty. The point $\tilde{H}(s_i,t_j)$ lies in exactly one sheet $V$ over $U_{ij}$. Since $C$ is connected and $\tilde{H}(C)\subseteq p^{-1}(U_{ij})$ is contained in the union of the disjoint open sheets, and $\tilde{H}{|}_C$ is continuous, the image $\tilde{H}(C)$ lies in the single sheet $V$: otherwise the preimages of two distinct sheets would disconnect $C$. Define on $R_{ij}$ $$\tilde{H}:=(p{|}_V{)}^{-1}\circ H.$$ On $C$ this agrees with the previously defined values, because both are points of $V$ with the same image under the injective $p{|}_V$. The sets $A_{ij}$ and $R_{ij}$ are closed in their union, so lemma 187.11 gives continuity on $A_{ij}\cup R_{ij}$, which is the next stage of the invariant. After the last rectangle the invariant reads: $\tilde{H}$ is defined and continuous on $I\times I$, lifts $H$, and restricts to $h_0$. ◻

Proof. Well-definedness of the product is exercise 187.5. Well-definedness of the inverse follows from it by reversing a homotopy in the first coordinate: if $H$ is a path homotopy from $\gamma$ to ${\gamma }^{\prime }$, then the map sending $(s,t)$ to $H(1-s,t)$ is a path homotopy from $\bar{\gamma }$ to the reversal of ${\gamma }^{\prime }$.

Associativity is exercise 187.6. For the unit, $c_{x_0}\cdot \gamma =\gamma \circ \phi$ with $\phi (s)=max(0,2s-1)$, and $\gamma =\gamma \circ \mathrm{id}$; the two reparametrizations agree at $0$ and $1$, so lemma 187.20 gives $c_{x_0}\cdot \gamma \simeq \gamma$ rel $\{0,1\}$, and symmetrically on the other side.

For inverses, define $K(s,t):=\gamma (min(2s,2-2s,1-t))$. This is continuous, being $\gamma$ applied to a minimum of three continuous functions with values in $[0,1]$. At $t=0$ it is $\gamma (min(2s,2-2s,1))=(\gamma \cdot \bar{\gamma })(s)$, since for $s\le \frac{1}{2}$ the minimum is $2s$ and for $s\ge \frac{1}{2}$ it is $2-2s$; at $t=1$ it is the constant $\gamma (0)=x_0$; and at $s\in \{0,1\}$ the first or second entry is $0$, so $K(0,t)=K(1,t)=x_0$. Hence $[\gamma ][\bar{\gamma }]=1$, and the same calculation with $\bar{\gamma }$ in place of $\gamma$ gives the other equation. ◻

Proof. (i) $f\circ (\gamma \cdot \delta )=(f\circ \gamma )\cdot (f\circ \delta )$ because concatenation is defined by cases on the parameter, which $f$ does not touch; composing a path homotopy with $f$ gives a path homotopy, so $f_{\ast }$ is well defined. The two remaining equations hold on representatives.

(ii) Well-definedness follows from exercise 187.5 applied twice. It is a homomorphism because $\eta \cdot \gamma \cdot \bar{\eta }\cdot \eta \cdot \delta \cdot \bar{\eta }\simeq \eta \cdot \gamma \cdot \delta \cdot \bar{\eta }$ by proposition 188.9, and ${\beta }_{\bar{\eta }}{\beta }_{\eta }[\gamma ]=[\bar{\eta }\cdot \eta \cdot \gamma \cdot \bar{\eta }\cdot \eta ]=[\gamma ]$ by the same calculation.

(iii) Let $\gamma$ be a loop at $x_0$ and consider $G(s,t):=H(\gamma (s),t)$, a homotopy from $f\circ \gamma$ to $g\circ \gamma$ which is not rel $\{0,1\}$: its two edges both traverse $\eta$. Define $$L(s,t):=(\eta {|}_{[0,t]}\cdot G(-,t)\cdot \overline{\eta {|}_{[0,t]}})(s),$$ where $\eta {|}_{[0,t]}(u):=\eta (tu)$. Each ingredient is continuous in $(s,t)$ jointly, and the three-fold concatenation is continuous by lemma 187.11. At $t=0$ this is $f\circ \gamma$ up to the reparametrization of lemma 187.20, at $t=1$ it is $\eta \cdot (g\circ \gamma )\cdot \bar{\eta }$, and both endpoints stay at $f(x_0)$. Hence $f_{\ast }[\gamma ]={\beta }_{\eta }(g_{\ast }[\gamma ])$. If $f$ is a homotopy equivalence with homotopy inverse $u$, then (i) and the displayed identity applied to $u\circ f\simeq \mathrm{id}$ and $f\circ u\simeq \mathrm{id}$ give that $f_{\ast }$ has a left and a right inverse up to the isomorphisms $\beta$, hence is an isomorphism. ◻

Proof. The value is an integer. $e(\tilde{\gamma }(1))=\gamma (1)=b$, and $e^{-1}(b)=\mathbb{Z}$ by example 188.3.

Independence of the representative. Let $H$ be a path homotopy from $\gamma$ to ${\gamma }^{\prime }$. By theorem 188.7 with $n=1$ there is a unique $\tilde{H}$ lifting $H$ with $\tilde{H}(-,0)=\tilde{\gamma }$. For fixed $t$, the map $s\mapsto \tilde{H}(s,t)$ lifts $H(-,t)$, which is a loop at $b$. The map $t\mapsto \tilde{H}(0,t)$ lifts the constant path at $b$ and starts at $0$, so by uniqueness in theorem 188.6 it is constant at $0$; therefore $\tilde{H}(-,t)$ is the lift starting at $0$, for every $t$. The map $t\mapsto \tilde{H}(1,t)$ is continuous with values in $e^{-1}(b)=\mathbb{Z}$, which is discrete by lemma 188.2; its domain $[0,1]$ is connected, so it is constant. Hence $\tilde{\gamma }(1)=\tilde{H}(1,0)=\tilde{H}(1,1)=\widetilde{{\gamma }^{\prime }}(1)$.

Homomorphism. Let $\gamma ,\delta$ be loops at $b$ with lifts $\tilde{\gamma },\tilde{\delta }$ starting at $0$, and put $m:=\tilde{\gamma }(1)$. Since $e(x+m)=e(x)$ for $m\in \mathbb{Z}$, the path $s\mapsto m+\tilde{\delta }(s)$ lifts $\delta$ and starts at $m$. So is continuous by lemma 187.11, lifts $\gamma \cdot \delta$, and starts at $0$; by uniqueness it is the lift, and its value at $1$ is $m+\tilde{\delta }(1)=\mathrm{deg}[\gamma ]+\mathrm{deg}[\delta ]$.

Surjectivity. The lift of ${\ell }_n$ starting at $0$ is $s\mapsto ns$, so $\mathrm{deg}[{\ell }_n]=n$.

Injectivity. Suppose $\mathrm{deg}[\gamma ]=0$, so $\tilde{\gamma }$ is a loop at $0$ in $\mathbb{R}$. Then $G(s,t):=(1-t)\tilde{\gamma }(s)$ is a path homotopy in $\mathbb{R}$ from $\tilde{\gamma }$ to the constant path at $0$, and $e\circ G$ is a path homotopy from $\gamma$ to $c_b$. Hence $[\gamma ]=1$. ◻

Proof. Theorem 187.26 makes the inclusion $S^1\hookrightarrow {\mathbb{R}}^2\setminus \{0\}$ a homotopy equivalence fixing $b$, so proposition 188.11(iii) makes the induced map an isomorphism, and theorem 188.12 identifies the source with $\mathbb{Z}$. ◻

Proof. A contractible space has trivial fundamental group by proposition 188.11(iii), since the one-point space does; but ${\pi }_1(S^1,b)\cong \mathbb{Z}$ is not trivial. ◻

Proof. (i) is theorem 188.7. For (ii), let $H:Z\times [0,1]\to X$ and $h_0=(H(-,0),g_0):Z\to X\times F$ be given; then $\tilde{H}(z,t):=(H(z,t),g_0(z))$ is continuous by lemma 187.7, lifts $H$, and restricts to $h_0$. The fiber over $x$ is $\{x\}\times F\approx F$. ◻

Proof. Let $H:I^n\times [0,1]\to X$ and $h_0:I^n\to X{\times }_{B}E$ be given with $q\circ h_0=H(-,0)$; write $h_0=(H(-,0),k_0)$, so that $p\circ k_0=f\circ H(-,0)$. Apply the homotopy lifting property of $p$ to the homotopy $f\circ H$ and the initial lift $k_0$, obtaining $\tilde{K}$ with $p\circ \tilde{K}=f\circ H$ and $\tilde{K}(-,0)=k_0$. Then $(H,\tilde{K})$ is continuous into $X\times E$ by lemma 187.7, lands in $X{\times }_{B}E$ by the displayed equation, and is continuous into that subspace by lemma 187.5(iii). Its value at time $0$ is $h_0$. Finally $q^{-1}(x)=\{x\}\times p^{-1}(f(x))$. ◻

Proof. Every argument of proposition 188.9 applies verbatim after replacing the single parameter $s$ by the $k$-th coordinate and carrying the remaining $n-1$ coordinates unchanged through every formula: the homotopies used there were built by reparametrizing one coordinate, and $f$ is constant at $x_0$ on every face, so each intermediate map still sends $\partial I^n$ to $x_0$. For instance the inverse homotopy is $K(x,t):=f(x_1,\dots ,min(2x_k,2-2x_k,1-t),\dots ,x_n)$, whose boundary values are computed exactly as before. ◻

Proof. Both sides of equation 188.1 are defined by the same case distinction on $x_j$ and $x_k$: on the region $x_k\le \frac{1}{2}$, $x_j\le \frac{1}{2}$ both are $f$ with the two coordinates doubled; on $x_k\ge \frac{1}{2}$, $x_j\le \frac{1}{2}$ both are $g$; on $x_k\le \frac{1}{2}$, $x_j\ge \frac{1}{2}$ both are $h$; and on the remaining region both are $l$. So they are equal as functions.

Write $u:={\cdot }_k$ and $v:={\cdot }_j$ for the induced operations on classes, and let $1$ be the class of the constant map, which is a unit for both by proposition 188.19. For classes $a,b$, $$aub=(av1)u(1vb)\stackrel{equation188.1}{=}(au1)v(1ub)=avb,$$ so the two operations agree; and $$aub=(1va)u(bv1)\stackrel{equation188.1}{=}(1ub)v(au1)=bva=bua.$$ ◻

Proof. Continuity and values. $2-x_n\ge 1$, so $u$ is continuous with values in $[\frac{1}{2},1]$, and $z(x)$ is the point where the segment from $c$ to $x$ meets the hyperplane $x_n=1$: its $n$-th coordinate is $2(1-u)+u{x}_n=2-u(2-x_n)=1$. Its first $n-1$ coordinates form a convex combination of $(\frac{1}{2},\dots ,\frac{1}{2})\in I^{n-1}$ and $x^{\prime }\in I^{n-1}$, hence lie in $I^{n-1}$. So $z(x)\in T$. For $x\in T$ we have $u(x)=1$ and $z(x)=x$; in particular $z$ is the identity on $\partial T$.

Injectivity on $J^{n-1}$. Suppose $x\ne y$ in $J^{n-1}$ with $z(x)=z(y)$. Then $x$ and $y$ lie on one ray from $c$, and the intersection of that ray with the convex set $I^n$ is a segment $S$ containing $x$, $y$, and $z(x)$. Along the ray, the $n$-th coordinate is strictly decreasing, since the direction $x-c$ has $n$-th component ; hence $z(x)$, having $n$-th coordinate $1$, is the endpoint of $S$ nearer $c$, and $x\ne y$ forces one of them, say $x$, to lie in the interior of $S$. A point in the interior of a segment contained in a convex set and lying on the boundary of that set forces the whole segment into a single face: if $x$ lies in the face $x_i=ϵ$ with $ϵ\in \{0,1\}$ and , then $S$ lies in that face, because $\lambda \mapsto \frac{1}{2}+\lambda (x_i-\frac{1}{2})$ is affine and leaves $[0,1]$ strictly on one side of $x$ unless it is constant; and if $x$ lies in the face $x_n=0$, then $x$ is the far endpoint of $S$, not an interior point. In the first case $z(x)$ also has $i$-th coordinate $ϵ$, so $z(x)\in \partial T$ and hence $z(x)=x$ by the previous paragraph applied to $x\in S\subseteq$ that face — but then $S$ degenerates to the point $x$, contradicting $x\ne y$.