Hurewicz Whitehead Freudenthal and Cw Approximation — Examples

If $n\ge 2$, apply [F4] with a singleton indexing set and its sphere based at $b$. This is the given sphere, not a wedge with an extra summand. It is a CW complex, is path connected, and has zero positive homotopy groups below $n$. Thus it satisfies exactly the $(n-1)$-connectivity hypothesis of [F1]. Assuming [A1], [F1] and [F2] give ${\pi }_n(S^n,b)\stackrel{h}{\to }{H}_n(S^n;\mathbb{Z})=\mathbb{Z}[S^n]$ as an isomorphism. This proves the asserted first nonzero group by the Hurewicz route.

For any based self-map $u$ of the oriented sphere, degree means the integer $d$ for which $u_{\ast }[S^n]=d[S^n]$, as used in [F3]. Consequently [F5] gives $h([u])=d[S^n]$. In particular $h([\mathrm{i}\mathrm{d}])=[S^n]$, and $h$ followed by the coefficient map $d[S^n]\mapsto d$ is exactly degree. This calculation also holds when $n=1$ because the defining formula in [F5] includes degree one.

For $n=1$, [F3] directly gives ${\pi }_1(S^1,b)\cong \mathbb{Z}$. There is no positive integer $i<1$, so the lower-vanishing assertion has no instance. Step 2.1 identifies the Hurewicz image of each degree class and in particular of the identity. For every $n\ge 2$, [F3] also supplies the degree isomorphism independently of [F1], while [F4] already gave lower vanishing choice-free. Thus the AC hypothesis belongs only to the indicated general-Hurewicz derivation, not to the independent sphere calculation.

The group in degree $n$ is nonzero because its identity-map class has coefficient $1$; the identity element of that group is instead the constant-map class with coefficient $0$. Negative coefficients are inverse classes by step 2.1. Empty spheres and $n=0$ are outside the hypothesis. Every specified basepoint is permitted by [F3] and the based singleton construction in [F4]; no family of basepoints or orientations is selected. This establishes the calculation and its orientation normalization in all stated degrees.

By [F2], $W$ has a CW structure with one vertex and one $n$-cell for each $j$, each attached by its constant boundary. It is path connected, and ${\pi }_i(W,b)=0$ for $0<i<n$. Thus $W$ is $(n-1)$-connected and meets [F1]'s CW and nonemptiness hypotheses. In particular $n=2$ gives simple connectivity, rather than assuming it from an unstated wedge principle. By [A1] and [F1], $h:{\pi }_n(W,b)\to H_n(W;\mathbb{Z})$ is an isomorphism.

Formula [F4] gives $h([{\iota }_j])=({\iota }_j{)}_{\ast }[S_j^n]=e_j\mathrm{.}$ Since $h$ is a homomorphism, it follows for every integer vector $a=(a_j{)}_{j\in J}$ that $h({\sum }_{j\in J}{a}_j[{\iota }_j])={\sum }_{j\in J}{a}_j{e}_j\mathrm{.}$ The sum on the left is well defined: $h$ is an injective homomorphism into the abelian homology group, so its source is abelian (the image of each commutator is zero, hence the commutator itself is the identity). Finite sums are therefore independent of the order, and negative coefficients mean inverse classes.

For $\alpha \in {\pi }_n(W,b)$ define $c_j(\alpha )$ by $(p_j{)}_{\ast }h(\alpha )=c_j(\alpha )[S_j^n]\mathrm{.}$ These integers are unique by [F3], and they form a vector in the finite direct sum because $J$ is finite. Let $T(a)={\sum }_j{a}_j[{\iota }_j]$. The two composites are identities: step 2.1 and the coordinate inverse of [F3] give $c(T(a))=a$; conversely [F3] expresses $h(\alpha )={\sum }_j{c}_j(\alpha )e_j=h(T(c(\alpha )))$, and injectivity of $h$ gives $T(c(\alpha ))=\alpha$. Both maps are homomorphisms by [F3], [F4] and finite additivity. This proves the stated isomorphism with precisely the inclusion basis, not just an abstract equality of ranks.

A singleton $J$ recovers one sphere and its identity generator. The zero vector corresponds to the constant class; negative vectors correspond to inverse classes by step 2.1. Empty $J$ is excluded in the stated example, although [F2] and [F3] consistently assign it a point and a zero positive group. The restriction $n\ge 2$ is required for [F1]'s higher Hurewicz isomorphism; no free-abelian claim for a wedge of circles is being made. The AC cost of this derivation is exactly [A1]; choosing an ordering of every finite set or a family of orientation representatives was not used. The claimed connectivity and both inverse formulas are now proved.

Apply [F1] to the empty fixed subcomplex to obtain a cellular map $g:X\to Y$ and a homotopy $E:f\simeq g$. Use its finite clause when $X,Y$ are finite; otherwise use [A1]. By the prism identity [F3], $g_{\ast }=f_{\ast }$ in each homology degree, so $g$ is also a homology equivalence. Form its ordinary CW cylinder $M$ with source $j:X\hookrightarrow M$, target $k:Y\hookrightarrow M$ and retraction $r:M\to Y$ in [F2]. Then $rj=g$, $rk=\mathrm{i}\mathrm{d}$ and $kr\simeq {\mathrm{i}\mathrm{d}}_M$. Thus $r_{\ast }$ in homology has inverse $k_{\ast }$ by [F3], and $j_{\ast }=r_{\ast }^{-1}{g}_{\ast }$ is an isomorphism in every degree.

For $i\ge 1$ consider $H_i(X)\stackrel{j_{\ast }}{\to }{H}_i(M)\to H_i(M,j(X))\stackrel{\delta }{\to }{H}_{i-1}(X)\stackrel{j_{\ast }}{\to }{H}_{i-1}(M)$. Since the rightmost map is injective, every relative class has zero boundary and comes from $H_i(M)$. Since the leftmost map is surjective, that entire image in the relative group is zero. Hence $H_i(M,j(X))=0$. In degree zero the relative group is the cokernel of the surjective $j_{\ast }:H_0(X)\to H_0(M)$, so it too is zero.

The space $M$ is path connected: each of its points has its cylinder track to $k(Y)$, and $Y$ is path connected. The retraction's all-basepoint homotopy isomorphisms [F2] show ${\pi }_1(M,j(x))=0$ for every $x\in X$, since $Y$ is simply connected. Both component sets of $j(X)$ and $M$ are singletons. The pointed tail in [F4] therefore shows that every relative degree-one class comes from ${\pi }_1(M,j(x))$ and is distinguished. Thus the CW pair $(M,j(X))$ is $1$-connected in the exact sense of [F4], and its subspace is nonempty and simply connected.

Induct on the integer $n\ge 2$. Suppose the pair is $(n-1)$-connected, starting with step 2.2. At each arbitrary $x\in X$, [F5] identifies ${\pi }_n(M,j(X),j(x))$ with $H_n(M,j(X))=0$ from step 2.1. Hence all these relative groups vanish, and the component condition is unchanged, so [F4] makes the pair $n$-connected. Induction proves vanishing in every positive relative degree at every $x$. This is induction on a property, not the selection of an infinite sequence of homotopies or inverse maps. In particular its use of [F5] is choice-free even though its all-data weak model can be infinite.

For every $i\ge 1$ the exact segment ${\pi }_{i+1}(M,j(X),j(x))\to {\pi }_i(X,x)\stackrel{j_{\ast }}{\to }{\pi }_i(M,j(x))\to {\pi }_i(M,j(X),j(x))$ has trivial outside terms by step 3.1. Exactness gives zero kernel and full image for the middle homomorphism. This also works for $i=1$, whose right outside object is pointed rather than a group. The component bijection was checked in step 2.2. Thus $j$ is weak by [F6]. The retraction in [F2] is weak at all basepoints, so $g=rj$ is weak, including its component map.

Apply Whitehead [F6] to $g:X\to Y$. Under [A1] its general clause applies. For finite $X,Y$, its finite clause applies and requires no choice; the preceding approximation used its finite clause and the Hurewicz comparison was choice-free. Let $h:Y\to X$ be the resulting homotopy inverse, so $hg\simeq {\mathrm{i}\mathrm{d}}_X$ and $gh\simeq {\mathrm{i}\mathrm{d}}_Y$. Composing $E$ with $h$ on its two sides gives $hf\simeq hg$ and $fh\simeq gh$. Concatenation gives both $hf\simeq {\mathrm{i}\mathrm{d}}_X$ and $fh\simeq {\mathrm{i}\mathrm{d}}_Y$, proving the conclusion for the original $f$. These are unbased homotopies, so the approximation never required an unstated fixed basepoint.

Empty endpoint complexes are excluded by the stated meaning of simply connected; a point endpoint or an identity map satisfies the same argument. All relative homology groups, including degree zero, were checked in step 2.1, and the first induction degree $n=2$ meets [F5]'s simple-connectivity hypothesis by step 2.2. No finite-dimensional upper bound is needed for the induction on group vanishing. The finite branch has only finite approximation and finite Whitehead choices; the arbitrary branch uses [A1] exactly in [F1] and [F6]. Both homotopy-inverse identities are established in step 5.1, with no conclusion asserted for non-simply-connected spaces.

For $n\ge 2$, [F3] supplies exactly the $(n-1)$-connectivity and CW hypotheses of [F1]. For $n=1$, the circle is the quotient of one closed interval with its endpoints identified, with one vertex and one open edge. Its characteristic interval is a quotient map, so this is its finite CW weak topology; path connectedness follows from its interval parametrization. Thus it is $0$-connected, which is all [F1] requires in this case. The constant loops and the sole component are included; no positive connectivity is claimed for $S^1$.

The map $\mathrm{\Phi }:\mathrm{\Sigma }{S}^n⟶S^{n+1},[x,t]⟼(\sqrt{1-t^2}x,t)$ is well defined and continuous by [F2]: at either endpoint the first coordinates vanish independently of $x$. It is bijective, since for $-1<t<1$ the inverse recovers $t$ as the last coordinate and $x$ by division by $\sqrt{1-t^2}$, while the two poles have precisely their respective apex preimages. Its source is compact as a quotient of the compact set in [F4], so [F4] makes it a homeomorphism. The lower apex goes to $(0,\dots ,0,-1)$. Postcomposition with this homeomorphism and its inverse gives inverse maps on based homotopy classes and preserves concatenation, so [F1] and step 1.1 give the asserted sphere ranges. If a different target basepoint is desired, a specified sphere path and [F5] transport these isomorphism or surjectivity assertions. At $t=\pm 1$ no division formula is used.

Put $i=n+k$ with $k\ge 0$. The inequality $i<2n-1$ is exactly $n+k<2n-1$, or $n>k+1$. If it holds, it continues to hold with $n$ replaced by $n+r$ for every $r\ge 0$, so every subsequent suspension transition is an isomorphism by step 2.1. Thus the sequence is constant up to these specified isomorphisms from that index onward. At equality $n=k+1$, $i=2n-1$ is precisely the surjective endpoint, with no injectivity conclusion supplied. For example $k=0$ is in the isomorphism range for $n\ge 2$ and only the surjective endpoint at $n=1$; $k=1$ is in the isomorphism range for $n\ge 3$ and at the endpoint for $n=2$. Degree zero consists of the singleton components and the trivial target fundamental group by [F1]. The case $n=0$ is outside this assertion. Neither these arithmetic bounds nor the compact quotient and basepoint comparisons introduce AC.

The map $f$ is continuous: the inverse image of every open subset of $Q$ is a subset of $Q_d$, hence open. Every continuous path $p:I\to Q$ is constant. Indeed, if two of its values differ, restrict to the closed interval between their parameters and compose with the continuous inclusion $Q\to \mathbb{R}$ from [F2]. By [F5] an irrational lies strictly between those values; [F4] supplies a parameter with that irrational value, impossible for a map into $Q$. The same assertion holds for paths into $Q_d$, since composing with $f$ gives a path into $Q$ and $f$ is the identity on the underlying set.

Every path component in either space is therefore a singleton. At each rational $q$, the map on components sends $\{q\}$ to $\{q\}$, so it is bijective. For an integer $k\ge 1$, let $a:I^k\to Q$ be a continuous based cube with boundary value $q$. Any $u,v\in I^k$ are joined by the continuous segment $t\mapsto (1-t)u+tv$, which remains in the cube coordinate by coordinate. Its composite with $a$ is a path, hence constant by step 1.1. Thus $a$ is constant, and its value is $q$ because the boundary is nonempty. For a cube into $Q_d$, compose with $f$ to reach the same conclusion. Each based homotopy group consequently has exactly one element in every positive degree, including degree one. The induced map is the unique homomorphism between trivial groups, hence an isomorphism. Together with the component calculation this proves that $f$ is a weak homotopy equivalence by [F1].

If $g:Q\to Q_d$ were a homotopy inverse, [F3] would give a homotopy $H:Q\times I\to Q$ between $fg$ and ${\mathrm{i}\mathrm{d}}_Q$. For each $q\in Q$, the function $t\mapsto H(q,t)$ is a continuous path, hence constant by step 1.1. Evaluating at the two endpoints gives $f(g(q))=q$. Since $f$ is the identity set map, $g(q)=q$ for every rational $q$. Thus any possible homotopy inverse is forced to be the identity set map $Q\to Q_d$.

That identity is not continuous. The singleton $\{0\}$ is open in $Q_d$, but is not open in $Q$: if $\{0\}=U\cap Q$ for an open real set $U$, then $0\in U$ supplies $\epsilon >0$ with $(-\epsilon ,\epsilon )\subset U$. By [F5] there is a rational $r\in (0,\epsilon )$, giving $r\in U\cap Q$ and $r\mathrm{\ne }0$, a contradiction. Hence the necessary $g$ of step 2.2 cannot be continuous, and $f$ is not a homotopy equivalence.

This is a nonempty example with infinitely many components; it does not assert that $Q$ is weakly contractible. The degree-zero condition is the bijection of those singleton components, while every positive group at every rational basepoint is zero. A constant cube, a constant homotopy and the interval endpoints all appear in steps 1.1–2.2; none is excluded. The argument instantiates an irrational or rational in one specified interval at a time and uses the choice-free intermediate value theorem, so no choice axiom is needed. Steps 2.1 and 3.1 give respectively the hypothesis and the failed conclusion of the asserted implication, completing the counterexample.

The wedge $W$ has one vertex and two open edges, and $r$ is a continuous loop by endpoint agreement and finite pasting. It is a cellular attaching map into that $1$-skeleton with finite support, so [F1] gives the stated finite CW space $X$, with $W$ a closed subcomplex and $\chi$ a homeomorphism on the open disk onto its cell. Every point of $W$ is connected to $v$ along its edge, and every interior-disk point is connected radially to a boundary point and then to $v$. Thus both $W$ and $X$ are path connected. The quotient $W\mathrm{/}{S}_a^1$ is $S_b^1$, with its usual quotient interval topology. The quotient $X\mathrm{/}W$ is $D^2\mathrm{/}{S}^1\cong S^2$: identify the open disk with ${\mathbb{R}}^2$ by $z\mapsto z\mathrm{/}(1-\mathrm{\mid }z\mathrm{\mid })$, follow by inverse stereographic projection, and send the collapsed boundary to infinity. Continuity at that point follows because escaping every bounded set corresponds to approaching the disk boundary; the same description supplies inverse continuity. These are the actual quotient identifications, by [F10], not merely homology equivalences inferred from a cellular complex.

Let $k:S_a^1\to W$ and $s:S_b^1\to W$ be inclusions, and $p_a,p_b$ the maps collapsing the other loop. They are continuous by the quotient interval test and satisfy $p_{a}k=1$, $p_{b}s=1$, with constant cross composites. The pair exact sequence [F3], the vanishing of positive point homology from [F4], and the $H_0$ isomorphism for the path-connected $S_b^1$ from [F9] show directly that $H_j(S_b^1)\to H_j(S_b^1,\ast )$ is an isomorphism for $j>0$. Thus [F2] and [F4] make $H_j(W,S_a^1)=H_j(S_b^1,\ast )$ zero for $j\ge 2$ and $\mathbb{Z}$ for $j=1$. The pair sequence [F3] gives $H_j(W)=0$ for $j\ge 2$. In degree one it gives $0\to \mathbb{Z}\stackrel{k_{\ast }}{\to }{H}_1(W)\stackrel{(p_b{)}_{\ast }}{\to }\mathbb{Z}\to 0$: the connecting arrow to $H_0(S_a^1)$ vanishes because its next map to $H_0(W)$ is an isomorphism by [F9]. Its right inverse is $s_{\ast }$. For any $z\in H_1(W)$, $z-s_{\ast }(p_b{)}_{\ast }z$ lies in the image of $k_{\ast }$, and applying $(p_a{)}_{\ast }$ gives its unique coefficient. Therefore $H_1(W)=\mathbb{Z}\alpha \oplus \mathbb{Z}\beta$, where $\alpha =a_{\ast }[S^1]$ and $\beta =b_{\ast }[S^1]$, with coordinate inverse $((p_a{)}_{\ast },(p_b{)}_{\ast })$. Also $H_0(W)=\mathbb{Z}$. This proves the wedge calculation directly without invoking the higher-sphere wedge lemma outside its $n\ge 2$ hypothesis.

Compute the fundamental group using an actual open cover. In the disk coordinates of $\chi$ put $U=X\setminus \chi (\{\mathrm{\mid }z\mathrm{\mid }\le 1\mathrm{/}3\})$ and $V=\chi (\{\mathrm{\mid }z\mathrm{\mid }<1\})$. The first is open by the quotient criterion: its inverse image is all of $W$ and the relatively open disk annulus $\mathrm{\mid }z\mathrm{\mid }>1\mathrm{/}3$. The second is open because its inverse image is the open disk and the empty subset of $W$. They cover $X$ and intersect in $1\mathrm{/}3<\mathrm{\mid }z\mathrm{\mid }<1$, an annulus. The radial homotopy $ru\mapsto ((1-t)r+t)u$ on the annular part of $U$, fixing $W$, descends by [F10] and retracts $U$ onto $W$. The disk $V$ is contractible and the annulus retracts onto a circle. All three spaces are path connected. Choose a basepoint in the annulus and the radial path from it to $v=r(1)$ on $W$; the retraction's basepoint track is this path, so [F8] supplies the based comparison. Its positive annulus generator retracts to the attaching word $a^2{b}^{-3}$, with this path fixing the whisker. Thus [F6, F7] give ${\pi }_1(X,v)\cong ⟨a,b\mid a^2{b}^{-3}=1⟩\mathrm{.}$ Indeed the pushout is the free group on $a,b$ modulo the normal closure of that word: a map from it to any group is exactly a choice of images of $a,b$ satisfying that relation, which is the van Kampen universal property for this cover.

By [F2], the pair $(X,W)$ has $H_2(X,W)=\mathbb{Z}$ and all other positive relative groups zero, using $X\mathrm{/}W=S^2$ and [F4]. The characteristic-disk map $\chi :(D^2,S^1)\to (X,W)$ induces an isomorphism in relative homology: the quotient square of [F2] identifies it with the homeomorphism $D^2\mathrm{/}{S}^1\to X\mathrm{/}W$ from step 1.1. The disk has no positive homology by [F11], so the pair sequence [F3] makes $\delta :H_2(D^2,S^1)\to H_1(S^1)$ an isomorphism. Let $[D^2,S^1]$ be the unique class mapping to the positive circle generator of [F4], and put $c={\chi }_{\ast }[D^2,S^1]$. Naturality of the boundary map [F3] gives $\delta c=r_{\ast }[S^1]$. By the degree-one additivity and inversion formula in [F5] and the actual word defining $r$, this is $2\alpha -3\beta$. Thus the pair sequence reduces in its only nonzero positive segment to $0⟶H_2(X)⟶\mathbb{Z}\stackrel{t\mapsto (2t,-3t)}{\to }{\mathbb{Z}}^2⟶H_1(X)⟶0.$ In degrees at least three its surrounding groups vanish, so $H_j(X)=0$ there. The displayed integer map is injective, since $2t=0$ implies $t=0$ in $\mathbb{Z}$, so $H_2(X)=0$ too.

The homomorphism $\lambda :{\mathbb{Z}}^2\to \mathbb{Z}$, $\lambda (u,v)=3u+2v$, kills $(2,-3)$ and sends $(1,-1)$ to $1$. It induces an isomorphism ${\mathbb{Z}}^2\mathrm{/}\mathbb{Z}(2,-3)\to \mathbb{Z}$: for every $(u,v)$, $(u,v)-(3u+2v)(1,-1)=-(u+v)(2,-3),$ so its inverse sends $k$ to $k[(1,-1)]$. The two formulas compose to the identity, proving both injectivity and surjectivity. Define $f:S^1\to X$ by first following $a$ and then $b^{-1}$, using the two half-intervals. By [F5], $f_{\ast }[S^1]$ has coordinates $(1,-1)$ modulo the attaching vector, so the just-proved isomorphism sends it to $1$. Hence $f_{\ast }$ is an isomorphism on $H_1$. In degree zero it takes a point to the sole component and is an isomorphism by [F9]. In every degree at least two both source and target groups vanish by [F4] and step 3.1. This verifies the promised integral homology equivalence in every degree, including its generator normalization.

Send $a$ to the transposition $A=(12)$ and $b$ to the cycle $B=(123)$, acting as permutations of $\{1,2,3\}$ with rightmost composition first. They obey $A^2=B^3=1$, so step 2.2's universal property gives a homomorphism from ${\pi }_1(X)$ to that permutation group. Its image is nonabelian: $(AB)(1)=1$, whereas $(BA)(1)=3$. In fact it is all $S_3$: $ABA=B^{-1}$, so the six distinct permutations $1,B,B^2,A,AB,A{B}^2$ comprise the subgroup generated by $A,B$ and exhaust the permutations of three letters. Therefore ${\pi }_1(X)$ is nonabelian, while ${\pi }_1(S^1)$ is abelian by [F7]. A homotopy equivalence would induce an isomorphism of these groups by [F8], impossible. Thus the particular $f$ in step 4.1 is not a homotopy equivalence despite all its integral homology isomorphisms.

This is a finite nonempty connected CW witness with exactly one vertex, two edges and one $2$-cell; no wild space or missing CW hypothesis is involved. The zero higher homology groups and the one-dimensional generator map were calculated, and the attaching vector has no kernel despite its mixed signs. The annular cover excludes the disk center but includes all boundary identifications in $U$; its homotopy fixes them, as required for quotient descent. The finite permutation calculation witnesses the failed fundamental-group conclusion. Only finitely many cells, maps and representatives are instantiated, and all supplier clauses used above are choice-free. No arbitrary-cell choice or general Hurewicz isomorphism is used.