Higher Homotopy Groups and Cofiber Sequences — Examples

Send [a] to $([p_{X}a],[p_{Y}a])$. Projecting a boundary-fixed homotopy gives boundary-fixed coordinate homotopies, so the map is well-defined. Conversely pair any representatives u,v to obtain $(u,v):I^n\to X\times Y$, continuous by F1 and constant at (x0,y0) on the boundary. Pairing the two homotopies proves independence of representatives. The composites are the identity because projections of (u,v) are u,v and pairing the projections of a recovers a pointwise.

Projection commutes with both half-cube formulas, hence the bijection is a homomorphism by F2. Two product points are joined by a path exactly when both coordinate pairs are joined: project a path in one direction and pair the two paths in the other using F1. This proves the component statement. In the displayed instance, substitution of 2t and 2t−1 gives exactly the two polynomial formulas; both equal (0,0) at their common endpoint t=1/2. The homotopy $(1-s)a(t)$ also contracts that particular loop, with both endpoints fixed.

The homotopy $H(x,t)=(1-t)x+t{x}_0$ stays in the disk by convexity, is continuous by F1 and fixes x0. Applying it to any based cubical representative proves every positive absolute group of $D^m$ trivial. Hence for k≥2 the exact segment $0\to {\pi }_k(D^m,S^{m-1})\stackrel{\mathrm{\partial }}{\to }{\pi }_{k-1}(S^{m-1})\to 0$ makes the boundary map an isomorphism.

For $2\le k<m$, apply F3 with source sphere dimension k−1 and target dimension m−1, using F5, to make the target of this isomorphism zero. For k=m, F4 identifies it with Z. Restricting the oriented characteristic map $\mathrm{i}\mathrm{d}:D^m\to D^m$ gives the oriented boundary identity, whose degree is +1. Thus its relative class is the asserted generator; for example when m=2 the boundary sends that disk to the once-traversed oriented circle, with coefficient 1.

For m≥2, F3 at source dimension zero makes $S^{m-1}$ path-connected. The low tail $0={\pi }_1(D^m)\to {\pi }_1(D^m,S^{m-1})\to {\pi }_0(S^{m-1})$ then makes the relative pointed set a singleton by F2. For m=1, write the disk as [-1,1] and x0=1 (reflection gives the other case). A representative α starts at a∈{−1,1} and ends at 1. The interpolation $(1-s)\alpha (t)+s((1-t)a+t)$ is continuous, stays in the interval, and fixes both endpoints. Thus all representatives with the same a are equivalent. A relative homotopy cannot change a, since a continuous path into the discrete two-point set is constant. Representatives $t\mapsto 1$ and $t\mapsto 2t-1$ therefore give exactly two classes.

The map $h([t])=e^{2\pi it}$ identifies the quotient circle with the unit circle: it is a continuous bijection, and inverse angular charts are continuous on arcs, including arcs crossing the quotient seam. Give it the counterclockwise orientation. Then $h{f}_d{h}^{-1}(z)=z^d$. For d>0 the fibre over 1 is $e^{2\pi ik\mathrm{/}d}$, 0≤k<d. In positive angular coordinates at each preimage and at 1, the map is u↦du. Interpolating the positive coefficient d to 1 on a sufficiently small arc gives a homotopy of punctured pairs. Thus F3 identifies its local multiplier with that of an orientation-preserving circle rotation, namely +1 by the identity and rotation homotopy, or the singleton-fibre case of F4. For d<0 the same calculation reduces to angular reflection, whose degree is −1 by the circle clause of F5. Hence F4 gives degree d for every nonzero d. For d=0 the map is constant and F5 gives degree zero.

For the fundamental group take U to be the target circle together with cone heights s<2/3, and V to be cone heights s>1/3 including its tip. They are open in the quotient, their union is the cone space, and their intersection is a circle times (1/3,2/3). U retracts to the target by decreasing height, V contracts to the tip by increasing height, and the intersection retracts to its circle. All are path-connected. Choose the basepoint at source [0], height 1/2 and transport to the target vertex along the height segment. The overlap generator maps to a^d in π1(U) by F9 and trivially in π1(V). Thus F8 gives the presentation $⟨a\mid a^d=1⟩=\mathbb{Z}\mathrm{/}d\mathbb{Z}$. This calculation does not infer H1 from π1.

The cone on the source circle is the disk via $[z,s]\mapsto (1-s)z$; its boundary at s=0 attaches by f_d. Thus the mapping cone has one vertex, one loop edge and one 2-cell. F2 and step 1.1 identify its attaching coefficient in F6 as d, so its cellular complex is $0\to \mathbb{Z}\stackrel{d}{\to }\mathbb{Z}\stackrel{0}{\to }\mathbb{Z}\to 0$. Direct kernels and images give the stated H0,H1,H2 and vanishing above dimension two, and F7 identifies them with singular homology.

When d=0 the cellular map is zero, so H1=H2=Z and the group relation is empty. When d=±1 the cellular map is an isomorphism and the group relation kills a, giving the claimed vanishings. For instance d=−2 gives H1=π1=Z/2Z and H2=0. No inference of contractibility from these invariants is made.

The inclusion U→U∨V is the pushout of the basepoint inclusion {* }→V along {* }→U. That inclusion is an unbased cofibration by well-pointedness, so F2 proves the claim. Collapsing U identifies the remaining V only at its existing basepoint, and compatible quotient maps in both directions give $(U\vee V)\mathrm{/}U\cong V$. F4 consequently identifies its cofiber with V up to based homotopy.

More explicitly that cofiber is $CU\vee V$: the original U is the base of the attached cone. On CU use $[u,s]\mapsto [u,s+t-st]$ and keep V fixed. At t=0 this is the identity, at t=1 all of CU is the tip, and the basepoint track is fixed. The quotient-times-I construction in F2 makes this a continuous based deformation onto V. The cofiber projection to ΣU collapses all of V, so its composite with the inclusion V→CU∨V is identically the basepoint. Under this explicit inverse of the equivalence, the connecting map is therefore constant. For U=V=S1 this gives the stated two-circle instance.

Let W be the wedge of two quotient circles, with vertex w and standard loops a,b. It has one vertex and two attached 1-cells, hence is a finite CW complex. By F1 the loop a:w×I→W extends to a homotopy $H:W\times I\to W$ with H(-,0)=id. Set g=H(-,1). Since a(0)=a(1)=w, both id and g are based; H is a free homotopy between them with basepoint track a.

F2 gives ${\mathrm{i}\mathrm{d}}_{\ast }={\beta }_a{g}_{\ast }$ and ${\beta }_a(c)=ac{a}^{-1}$. Applying it to b and multiplying in the free group yields $g_{\ast }(b)=a^{-1}ba$. By F3 and F4 this is the reduced three-letter word a inverse, b, a; it has no adjacent cancellation and differs from the one-letter reduced word b. Therefore $g_{\ast }\mathrm{\ne }{\mathrm{i}\mathrm{d}}_{\ast }$, although the maps are freely homotopic.

Take $X=\{0\}\cup \{1\mathrm{/}n:n\ge 1\}\subset \mathbb{R}$ and A={0}. The complement of A is open in X, since each point 1/n is isolated, so A is closed. Every path in X is constant: if two of its values differ, an irrational strictly between them would be a value of the composite real-valued path on the subinterval between those times by F3, but X contains only rational numbers.

If A→X had HEP, use target $S=X\times \{0\}\cup \{0\}\times I$, initial map x↦(x,0), and homotopy (0,t)↦(0,t). F1 would give $R:X\times I\to S$ fixing those prescribed points. Its first coordinate along {1/n}×I is a path in X starting at 1/n and therefore constant by step 1.1. The only point of S with first coordinate 1/n is (1/n,0); hence $R(1\mathrm{/}n,1)=(1\mathrm{/}n,0)$ for every n.

But $(1\mathrm{/}n,1)\to (0,1)$ in $X\times I$. Continuity would make the second coordinates of their R-images converge to the second coordinate of $R(0,1)=(0,1)$, namely 1. They are all zero by step 2.1, a contradiction. Thus this explicit closed inclusion is not a cofibration.