Higher Homotopy Groups and Cofiber Sequences

Inverse images commute with arbitrary intersections and finite unions, and send $\mathrm{\varnothing },X$ to $\mathrm{\varnothing },K$. Thus the k-closed sets are closed sets of a topology containing all original closed sets. Each original test $u:K\to X$ is continuous into $kX$ by that definition; the reverse follows by composing with the continuous identity $kX\to X$. Since the tests are identical, $k(kX)=kX$.

For a finite disjoint union, a k-closed subset restricts to a closed subset of each CG summand by testing the inclusion composed with every test; it is therefore closed. For $A$ closed in CG $X$ and $F$ k-closed in $A$, a test $u:K\to X$ restricts on the compact Hausdorff closed set $u^{-1}A$. Thus $u^{-1}F$ is closed there, hence in $K$. Consequently $F$ is closed in $X$, proving that the subspace $A$ is CG.

If $T$ is CG and $f:T\to X$ is continuous, then for every k-closed $F\subseteq X$ and test $v:K\to T$, $(fv{)}^{-1}F$ is closed. Hence $f^{-1}F$ is k-closed in $T$, thus closed. This proves continuity into $kX$; composition with $kX\to X$ proves the converse. Applied to $kT$, this also proves functoriality. A compact Hausdorff $K$ is CG since its identity is a test.

Let $F$ be k-closed in the ordinary $X\times I$ and $(x,t)\mathrm{\notin }F$. The vertical test $s\mapsto (x,s)$ shows $F_x$ closed. Choose a closed interval neighbourhood $J$ of $t$ in $I$ disjoint from $F_x$, using relative intervals at 0 and 1. Set $V=\{y:(\{y\}\times J)\cap F=\mathrm{\varnothing }\}$. For a test $u:K\to X$, the inverse image of $F$ in the compact Hausdorff $K\times J$ is closed, hence compact. Its projection is compact and closed in $K$ and equals $K\setminus u^{-1}V$. Thus $V$ is k-open and hence open. The rectangle $V\times {\mathrm{int}}_{I}J$ misses $F$, so $F$ is ordinary closed. Therefore $X\times I$ is CG.

A family of continuous coordinates from CG $T$ induces a continuous map into the ordinary product, which lifts to its kification by step 2.1. Conversely projections from the k-product are continuous. Coordinate uniqueness proves the product property, and inverse coordinate rearrangements prove finite associativity and symmetry. For an ordinary quotient $q:X\to Q$ with $X$ CG, step 2.1 makes $q:X\to kQ$ continuous. Each k-closed $F\subseteq Q$ therefore has closed $q^{-1}F$, so quotient finality makes $F$ closed in $Q$.

The inverse image of a closed factor under a projection from a k-product is closed, hence CG by step 1.2. Its ordinary subspace topology identifies with the corresponding k-product: the continuous coordinate map in one direction comes from step 3.1, while its inverse is continuous into the subspace because its composite into the ambient product is continuous. Iterating handles products of closed inclusions. A compact test into a finite clopen decomposition splits into compact Hausdorff clopen domains. Testing each piece proves that kification commutes with that decomposition.

Cubes and their cylinders are compact Hausdorff; the zero-fold cube is a point. Step 1.1 therefore preserves all maps and homotopies from these domains. The underlying functions do not change, so all specified boundary equalities are preserved as well. Empty spaces and empty coproducts satisfy the same closed-set tests vacuously.

For a compact Hausdorff test $(v,w):K\to Y{\times }_{k}C(Y,Z)$, consider $a\mapsto w(a)(v(a))$. If its value at $a$ lies in open $O$, regularity gives a closed neighbourhood $L$ of $a$ with $w(a)(v(L))\subseteq O$. The set $N=w^{-1}W(v{\mathrm{\mid }}_L,L,O)$ is open and contains $a$. For $b\in N\cap \mathrm{int}L$, one has $w(b)(v(b))\in O$. Thus every test composite of evaluation is continuous, so evaluation $Y{\times }_{k}C(Y,Z)\to Z$ is continuous.

Given continuous $g:X{\times }_{k}Y\to Z$, each function $f(x):y\mapsto g(x,y)$ is continuous by the slice map. For tests $v:L\to X$, $u:K\to Y$, the map $(b,a)\mapsto g(v(b),u(a))$ is continuous on the compact Hausdorff product. The tube lemma says that the set of $b$ for which all its values on $\{b\}\times K$ lie in $O$ is open. This is $(fv{)}^{-1}W(u,K,O)$. Testing on $L$ proves continuity $X\to C_0(Y,Z)$, and the CG-source property lifts it to $C(Y,Z)$.

Conversely, a continuous $f:X\to C(Y,Z)$ uncurries continuously by composing its product with the evaluation of step 1.1. The two constructions are inverse since both give the same value $f(x)(y)$ at every pair. Precomposition and postcomposition preserve this equality, proving naturality.

The map $(h,x,y)\mapsto h(x)(y)$ is a composite of two continuous evaluations. Twice applying the map correspondence of steps 1.2–2.1 makes the induced map $C(X,C(Y,Z))\to C(X{\times }_{k}Y,Z)$ continuous. Conversely, start with evaluation $(h,x,y)\mapsto h(x,y)$ and curry successively in $y$ and $x$. This gives the inverse continuous map. Product reassociations are homeomorphisms by F1, so the displayed bijection is a homeomorphism.

Let $P$ be the ordinary quotient of $X{\times }_{k}Y$ by the stated relation and $p$ its quotient map. It is CG. The map $q\times \mathrm{i}\mathrm{d}$ descends to a continuous bijection $a:P\to Q{\times }_{k}Y$. The transpose $x\mapsto (y\mapsto p(x,y))$ is constant on each q-fibre and hence descends continuously to $Q\to C(Y,P)$. Uncurrying gives $b:Q{\times }_{k}Y\to P$ with $b(q(x),y)=p(x,y)$. Surjectivity of q and p gives $ab=\mathrm{i}\mathrm{d}$ and $ba=\mathrm{i}\mathrm{d}$. Thus $q\times \mathrm{i}\mathrm{d}$ is quotient.

For quotient maps $q:X\to Q$ and $r:Y\to R$, factor $q\times r=({\mathrm{i}\mathrm{d}}_Q\times r)(q\times {\mathrm{i}\mathrm{d}}_Y)$. Each factor is quotient by step 3.2 and symmetry, and their composite is quotient. Empty factors give empty products and the same inverse identities; no representative of a quotient fibre has been selected.

If $X$ is WH, its points are closed by singleton tests. For $u:K\to X$ compact Hausdorff, $L=u(K)$ is closed and compact. The surjection $u:K\to L$ is closed: any closed subset of $K$ is compact Hausdorff and its image is closed by WH. For distinct $a,b\in L$, the disjoint closed fibres can be separated by disjoint open $U,V\subseteq K$. The sets $L\setminus u(K\setminus U)$ and $L\setminus u(K\setminus V)$ are disjoint open neighbourhoods of $a,b$. Therefore $L$ is Hausdorff.

Conversely suppose the diagonal is closed. For tests $u:K\to X$ and $v:L\to X$, the set $E=\{(a,b):u(a)=v(b)\}$ is closed in the compact Hausdorff $K\times L$, hence compact. Its projection to $L$ is compact and closed and is exactly $v^{-1}(u(K))$. Thus $u(K)$ is k-closed in $X$ and therefore closed. This proves WH.

A test into a subspace composed with its inclusion has image closed in the ambient WH space, hence closed in the subspace. Kification preserves tests and has finer topology, so it preserves WH. Step 1.1 also shows that each test-image mapping subbasic open is a compact-Hausdorff-subspace subbasic open; the converse uses the inclusion as test.

If CG $X$ is WH, test its diagonal by $(v,w):K\to X{\times }_{k}X$. At $a$ with $v(a)\mathrm{\ne }w(a)$, regularity provides a closed neighbourhood $L$ of $a$ inside $v^{-1}(X\setminus \{w(a)\})$. Then $v(L)$ is closed by WH, and $\mathrm{int}L\cap w^{-1}(X\setminus v(L))$ is a neighbourhood of $a$ on which $v\mathrm{\ne }w$. Each test equality set is closed; compact generation makes the diagonal closed.

The diagonal of a finite k-product is the intersection of the inverse images of the factor diagonals. Hence it is closed and the product is WH. A compact test into a finite coproduct has compact Hausdorff clopen inverse-image pieces; their images are closed in their summands, so their finite union is closed in the coproduct. Closed subspaces are CG by F2 and WH by step 2.1. These prove the asserted finite closure properties, including the empty product and coproduct.

For $q:X\to Q$, the quotient $Q$ is CG and $q\times q$ is quotient. Therefore its diagonal is closed exactly when its inverse image, the stated fibre equivalence relation, is closed. Steps 2.2–1.2 identify this condition with WH of $Q$.

Each point evaluation $e_x:C(X,Y)\to Y$ is continuous since inverse images of opens are singleton-test subbasic opens. The diagonal of $C(X,Y)$ is ${\bigcap }_{x\in X}(e_x\times e_x{)}^{-1}{\mathrm{\Delta }}_Y$, hence closed. This mapping space is CG by definition, so is WH. Requiring a basepoint or either interval endpoint to map to the closed point $y_0$ cuts out a closed subspace. Such subspaces are CGWH by step 3.1. The empty intersection when $X$ is empty gives the one-point mapping space.

The compact-test and relative-homotopy assertions are F1. Its ordinary cylinder is CG and agrees with the k-product; F2 gives WH of that product. Thus it is CGWH.

Put $S=X⨿Y$, $P=(X\setminus A)⨿Y$ as a set, and let $q:S\to P$ be identity off $A$ and equal to $f$ on $A$. Give $P$ the quotient topology. In the four clopen pieces of $S{\times }_{k}S$, its equivalence relation is respectively ${\mathrm{\Delta }}_X\cup E_f$, $G_f$, $G_f^{\mathrm{o}\mathrm{p}}$, and ${\mathrm{\Delta }}_Y$, where $E_f=\{(a,a^{\mathrm{\prime }}):f(a)=f(a^{\mathrm{\prime }})\}$ and $G_f=\{(a,y):f(a)=y\}$. These formulas include every fibre: only points of $A$ are identified with points of $Y$, and two such points are equivalent exactly when their f-values agree.

The sets $E_f$ and $G_f$ are inverse images of ${\mathrm{\Delta }}_Y$ in $A{\times }_{k}A$ and $A{\times }_{k}Y$. They are closed there by F2, hence in the corresponding products with $X$ by F1. The two diagonals are closed. The relation in step 1.2 is therefore closed, and F2 proves $P$ CGWH. Maps from $S$ constant on that relation are precisely compatible maps from $X,Y$, so F4 proves the pushout property in Top and in CGWH.

The map $j:Y\to P$ is injective. For closed $F\subseteq Y$, one has $q^{-1}(j(F))=f^{-1}(F)⨿F$, which is closed in $S$ since $A$ is closed. Hence $j(F)$ is closed in $P$. This proves that $j$ is a closed embedding. Set-theoretically $X{\times }_{P}Y$ consists exactly of $(a,f(a))$ for $a\in A$. A continuous compatible pair from any space has X-component landing in the ordinary subspace $A$ and hence factors continuously there. This proves the pullback property. A closed $D\subseteq X$ disjoint from $A$ similarly embeds as a closed subspace, since it and each of its closed subsets are saturated for q.

The attaching subspace $X\times \{0\}$ is closed, as are a WH basepoint track and finite unions of such tracks with a cone end. Collapsing any such closed subspace is the preceding pushout with a point. Thus the resulting cylinder, reduced-cylinder, cone and suspension quotients are CGWH. Their quotient products use F3 and retain the fibre coordinate; no whole product subspace is inadvertently collapsed. When $A=\mathrm{\varnothing }$, the formula is simply the disjoint union, and when $A=X$, it is $Y$.

Evaluation $C_0(I,Y)\times I\to Y$ is continuous: at $(f,t)$ with $f(t)\in O$ open, choose a closed interval neighbourhood $J$ of $t$ contained in $f^{-1}O$. Then $W(J,O)\times {\mathrm{int}}_{I}J$ is a neighbourhood mapping into $O$. Relative interval neighbourhoods work at 0 and 1.

For continuous $H:X\times I\to Y$, the inverse image of $W(K,O)$ under $h(x)=H(x,-)$ is open: for any of its points $x$, the open set $H^{-1}O$ contains $\{x\}\times K$ and F1 supplies a tube. Thus h is continuous into $C_0(I,Y)$. Conversely compose $h\times {\mathrm{i}\mathrm{d}}_I$ with step 1.1. The resulting functions are inverse under evaluation at each $(x,t)$.

If a continuous $H:X\times I\to Y$ is constant on each fibre of $q\times {\mathrm{i}\mathrm{d}}_I$, its transpose is constant on q-fibres. By F2 it descends to a continuous $Q\to C_0(I,Y)$, which uncurries by step 2.1 to a continuous $Q\times I\to Y$. Apply this to the characteristic function of $U\subseteq Q\times I$ into the Sierpinski space with opens $\mathrm{\varnothing },\{1\},\{0,1\}$. Continuity of the characteristic function is exactly openness of U. If $(q\times \mathrm{i}\mathrm{d}{)}^{-1}U$ is open, the preceding descent makes U open. The map is surjective and continuous; hence it is quotient.

For CG spaces the kified correspondence is F4; the interval product already has its ordinary topology. Endpoint and basepoint equations are preserved pointwise by transpose and inverse transpose. Continuous maps from CG parameters landing in the corresponding subspace lift to its kification by the CG-source property contained in the conventions of F4. For WH targets these subspaces are closed and CGWH by F5; F3 supplies the closed-track quotient constructions. Applying the same correspondence to a cylinder parameter, or using the quotient-times-I conclusion of step 3.1, carries relative and based homotopies to relative and based homotopies.

The two affine maps $(s,u)\mapsto (2s,u)$ and $(s,u)\mapsto (2s-1,u)$ are continuous on the closed half-cubes. On their common face the values of $a,b$ are both $x_0$. Thus the concatenation is continuous by closed pasting. Its outer boundary maps to $x_0$, since either s is an endpoint or a coordinate of u is an endpoint.

If $A,B:I^n\times I\to X$ are boundary-fixed homotopies between the two respective pairs of representatives, paste $A(2s,u,t)$ and $B(2s-1,u,t)$. The seam values are $x_0$ for every t; the same boundary calculation applies. This is a continuous boundary-fixed homotopy between the concatenations. Permuting the selected coordinate with coordinate 1 gives the identical proof whenever its two faces are fixed.

For any continuous $\varphi :I\to I$ fixing endpoints, $a((1-t)s+t\varphi (s),u)$ is a boundary-fixed homotopy from a to its reparametrization. The coordinate formula is jointly continuous, not merely continuous separately in u. Taking ${\varphi }_R(s)=\min (2s,1)$ and ${\varphi }_L(s)=\max (2s-1,0)$ gives $a\ast e\simeq a\simeq e\ast a$. These are the loop-law formulas of F2 with u retained as a parameter.

For $w=(a\ast b)\ast c$, set $\varphi (s)=s\mathrm{/}2$ on $[0,1\mathrm{/}2]$, $s-1\mathrm{/}4$ on $[1\mathrm{/}2,3\mathrm{/}4]$, and $2s-1$ on $[3\mathrm{/}4,1]$. The pieces agree and fix endpoints. Substitution gives $w(\varphi (s),u)=(a\ast (b\ast c))(s,u)$ on all three intervals. Step 1.1 therefore proves associativity on classes.

The map equal to $a(2s(1-t),u)$ for $s\le 1\mathrm{/}2$ and $a(2(1-s)(1-t),u)$ for $s\ge 1\mathrm{/}2$ pastes continuously. It fixes the exterior boundary, begins at $a\ast a^{-}$ and ends at e. Applying the same formula to $a^{-}$ contracts $a^{-}\ast a$. Together with steps 1.1–2.1 and F1 this verifies the group axioms of F3.

For $n\ge 2$ let $\ast$ and $\circ$ concatenate in coordinates 1 and 2. Each has the same two-sided unit by step 1.1. Pasting four quarter-cubes gives $(a\circ b)\ast (c\circ d)=(a\ast c)\circ (b\ast d)$ on representatives. Therefore on classes $a\ast b=(a\circ e)\ast (e\circ b)=(a\ast e)\circ (e\ast b)=a\circ b$, whereas $a\ast b=(e\circ a)\ast (b\circ e)=(e\ast b)\circ (a\ast e)=b\circ a$. Hence the common operation commutes. For n=1 there is no second coordinate, and no commutativity claim is made.

After centering and doubling the cube, its interior is $(-1,1{)}^n$. The coordinate map $v_i\mapsto v_i\mathrm{/}(1-\mathrm{\mid }{v}_i\mathrm{\mid })$ is a homeomorphism to ${\mathbb{R}}^n$, with inverse $w_i\mapsto w_i\mathrm{/}(1+\mathrm{\mid }{w}_i\mathrm{\mid })$. Approaching the cube boundary sends the Euclidean norm to infinity, and conversely bounded images stay away from that boundary. Thus the map extends to a homeomorphism of the collapsed-boundary cube with ${\mathbb{R}}^n\cup \{\mathrm{\infty }\}$. Inverse stereographic projection $w\mapsto (2w\mathrm{/}(1+\mathrm{\parallel }w{\mathrm{\parallel }}^2),(\mathrm{\parallel }w{\mathrm{\parallel }}^2-1)\mathrm{/}(1+\mathrm{\parallel }w{\mathrm{\parallel }}^2))$ identifies this compactification with $S^n$, taking the quotient point $[\mathrm{\partial }{I}^n]$ to the north pole. Take that north pole as $s_0$ and choose the sphere orientation to agree with the cube interior; this gives the required based $h$.

Every boundary-constant map descends uniquely by F2, and every based spherical map pulls back to a cubical map. For a boundary-fixed homotopy, F3 makes its descent across $(I^n\to I^n\mathrm{/}\mathrm{\partial }{I}^n)\times {\mathrm{i}\mathrm{d}}_I$ continuous. Pullback is the inverse and preserves endpoint maps. The bijections on maps therefore induce inverse bijections on homotopy classes.

Collapse also the middle face $s_1=1\mathrm{/}2$. Each resulting half-cube with its boundary collapsed is an oriented based copy of the same sphere, using the positive affine rescalings $2s_1$ and $2s_1-1$ and the based homeomorphism $h$. The composite of this transported pinch with $a$ on the first copy and $b$ on the second pulls back to the defining formula for $a\ast b$. Hence the operations agree.

Postcomposition with f sends boundary-fixed homotopies to boundary-fixed homotopies, now with value $y_0$, by F2. Hence $f_{\ast }[a]$ is well-defined. Pointwise on both half-cubes, $f\circ (a\ast b)=(f\circ a)\ast (f\circ b)$, so F1 makes it a homomorphism. A path is similarly sent to a path, giving a well-defined map on components.

For based maps f,g one has $g\circ (f\circ a)=(g\circ f)\circ a$ and $\mathrm{i}\mathrm{d}\circ a=a$, proving functoriality. If $H:X\times I\to Y$ is a based homotopy, $H(a(u),t)$ is continuous by F2 and equals $y_0$ on the cube boundary for every t. Thus its two endpoints define the same class, giving $f_{\ast }=g_{\ast }$. The same formula on points gives equality on components.

If based maps f and h are inverse up to based homotopy, step 2.1 gives $h_{\ast }{f}_{\ast }=(hf{)}_{\ast }=\mathrm{i}\mathrm{d}$ and $f_{\ast }{h}_{\ast }=(fh{)}_{\ast }=\mathrm{i}\mathrm{d}$. Thus these are inverse homomorphisms for positive degrees and inverse pointed bijections in degree zero.

Write cube coordinates $(u,t)\in I^{n-1}\times I$. On the complement of J, send each $u_i\in (0,1)$ homeomorphically to $v_i=(2u_i-1)\mathrm{/}(1-\mathrm{\mid }2u_i-1\mathrm{\mid })$ and send $t\in [0,1)$ to $v_n=t\mathrm{/}(1-t)$. This identifies that complement with the closed upper half-space in ${\mathbb{R}}^n$. Approaching J is precisely escaping every bounded subset. The inverse stereographic map $v\mapsto (2v\mathrm{/}(1+\mathrm{\parallel }v{\mathrm{\parallel }}^2),(\mathrm{\parallel }v{\mathrm{\parallel }}^2-1)\mathrm{/}(1+\mathrm{\parallel }v{\mathrm{\parallel }}^2))$ therefore extends over J collapsed to the north pole. Its image is the closed hemisphere where coordinate n is nonnegative; projection dropping coordinate n identifies that hemisphere homeomorphically with a disk, with inverse inserting the nonnegative square root. Its boundary comes from F and its marked boundary point from J. For n=1 this is the compactified half-line, an interval.

Quotient descent and pullback identify representatives in the two models. The same holds for homotopies because the quotient times I is quotient. In particular a relative nullhomotopy in the disk model is $H:D^n\times I\to X$ with $H(z,0)=f(z)$, $H(z,1)=x_0$, $H(S^{n-1}\times I)\subseteq A$, and $H(b,t)=x_0$.

For $z\in D^n$ put $r=\mathrm{\parallel }z\mathrm{\parallel }$, $v(z)=z\mathrm{/}\max (1\mathrm{/}2,r)$ and $h(z)=\min (1,2-2r)$. Both are continuous, $v(z)\in D^n$ and $h(z)\in I$. The map $R_s(z)=((1-s)z+sv(z),sh(z))$ lies in $D^n\times I$ and fixes every rim point $(z,0)$ with r=1. At s=0 it is the bottom disk. At s=1, points with $r\le 1\mathrm{/}2$ lie in the top disk and points with $r\ge 1\mathrm{/}2$ lie in the side boundary. Thus $H(R_s(z))$ is a homotopy fixed on the whole boundary from f to a map into A. The formula has no singularity at z=0 and agrees on r=1/2.

Conversely suppose a boundary-fixed homotopy joins f to $g:D^n\to A$. The formula $g((1-s)z+sb)$ contracts g to $g(b)=x_0$ through maps into A fixing b, because the disk is convex. Concatenating this with the given homotopy produces a relative nullhomotopy. The two constructions prove both implications, including n=1, where the rim has two points.

For n≥2, concatenate and reverse in coordinate 1. Its two faces are part of J; hence F2 makes the product and pasted representative homotopies continuous. The reparametrizations and reversal contractions in F3 act only on coordinate 1. Every J-face remains at x0, and the last-coordinate-zero face continues to map into A. Thus those same explicit homotopies prove associativity, unit and inverses in the relative set.

If n≥3, coordinates 1 and 2 are both available without changing the distinguished coordinate. The four-quarter identity and the two-unit calculation of F3 therefore apply to relative classes, proving commutativity. If n=2 only one coordinate is available, and if n=1 none is; the argument makes no stronger claim in those degrees.

Restriction to F sends a relative homotopy to a boundary-fixed homotopy in A, since $\mathrm{\partial }F\subseteq J$. For n≥2 it commutes pointwise with coordinate-1 concatenation, so $\mathrm{\partial }[a\ast b]=\mathrm{\partial }[a]\ast \mathrm{\partial }[b]$. For n=1 a relative homotopy moves the initial endpoint along a path in A, so its component is well-defined. Constant representatives map to the distinguished element in every degree.

For a map of based pairs $\varphi :(X,A,x_0)\to (Y,B,y_0)$, composing a representative or its homotopy with φ preserves all triple conditions. Composition and identity act pointwise, and composition commutes with products and with restriction to F. A based pair homotopy $\mathrm{\Phi }$ gives the representative homotopy $(u,t)\mapsto \mathrm{\Phi }(a(u),t)$; it sends F into B and J to y0. This proves all functoriality and homotopy assertions.

At ${\pi }_n(X)$, an absolute class represented by a map into A is relatively null: in the disk model contract its domain to the marked boundary point, through maps into A. Conversely a class killed by j compresses into A with its full boundary fixed by F2; since that boundary was constant x0, the compressed representative defines an absolute class in ${\pi }_n(A)$ mapping to the original. This proves both image inclusions for all n≥1.

At ${\pi }_n(X,A)$ for n≥2, the boundary of an absolute representative is constant, so $\mathrm{\partial }{j}_{\ast }=0$. If $a(u,t)$ has distinguished face $h(u)=a(u,0)$ nullhomotopic in A, take a boundary-fixed $B(u,s)$ from h to x0. For a collar width $0<\lambda \le 1$ define $a_{\lambda }(u,t)=B(u,\lambda -2t)$ for $0\le t\le \lambda \mathrm{/}2$, and $a(u,(t-\lambda \mathrm{/}2)\mathrm{/}(1-\lambda \mathrm{/}2))$ for $\lambda \mathrm{/}2\le t\le 1$. At the seam both values are h. As λ goes from 0 to 1, using this formula for $0<\lambda \le 1$ and $a_0=a$, it gives a relative homotopy: near λ=t=0 both arguments tend to the common value h. At λ=1 the bottom value is $B(u,1)=x_0$, and all other faces are fixed, so it is an absolute representative.

At ${\pi }_n(A)$ for n≥1, a relative $(n+1)$-cube is itself an X-nullhomotopy of its distinguished face, so $i_{\ast }\mathrm{\partial }=0$. Conversely, if an A-based cube is nullhomotopic in X rel its boundary, that nullhomotopy, with time as the final coordinate, is a relative $(n+1)$-cube whose boundary is the given cube. This proves equality of kernel and image there.

At ${\pi }_1(X,A)$, a path α starts at some $a\in A$ and ends at x0. Its boundary component is distinguished precisely when a can be joined to x0 in A. Given a path $p:x_0\to a$ in A, the based loop $p\ast \alpha$ has the same relative class as α: attach the terminal segment $p{\mathrm{\mid }}_{[1-s,1]}$ ahead of α with width s/2. At s=0 it is α; at s=1 it is the loop, the changing initial endpoint stays in A, and the seam agrees at a. Conversely the initial point of a based loop is x0, and any relative homotopy keeps its initial endpoint within the same A-component.

At ${\pi }_0(A)$, the component of a maps to the distinguished X-component exactly when a path in X joins a to x0. Such a path is a relative degree-one representative with boundary component [a]. Conversely every relative path provides that connection. Components of X not meeting A are not constrained by this calculation.

Composition with a map of based pairs commutes pointwise with inclusion and distinguished-face restriction. Hence every square of the displayed sequence commutes by F1 and F3. Steps 1.1–1.5 establish exactness at all eligible terms, with the pointed-set interpretation in the low tail.

The attachment $A\times \{0\}\hookrightarrow A\times I$ is closed, so R is CGWH by F2. Use R as target in HEP, with the initial copy of X and the homotopy $a,t\mapsto [a,t]$. The required extension s satisfies $sc=\mathrm{i}\mathrm{d}$ on each summand, hence on R. Conversely, compatible data $f,h$ induce a continuous $d:R\to Z$ by F4, and $ds$ is the required extension.

A left inverse s makes c injective and gives a continuous inverse from its image. Moreover $c(R)=\{z:cs(z)=z\}$, closed by F3; hence c is a closed embedding. The free end $A\times \{1\}$ embeds closed in R: it is disjoint from the attaching end and every closed subset of it has closed saturated image in the quotient. Its image under c is $i(A)\times \{1\}$. Restricting the resulting closed embedding to the endpoint identifies i as a closed embedding.

When i is a closed inclusion, both $X\times \{0\}$ and $A\times I$ are closed in the cylinder. A function out of their union is continuous exactly when its restrictions are continuous and agree on the overlap, by finite closed pasting. Thus its subspace topology is the pushout topology of R, and c is the inclusion of that strip. The first equivalence becomes exactly the ordinary strip-retraction criterion.

For based HEP replace R and $X\times I$ by their reduced versions, collapsing the closed basepoint tracks. They are CGWH by F2. The same universal test uses based maps and yields a left inverse; conversely the composite extension is based. Thus c is again a closed embedding by the equalizer argument. In each reduced space an endpoint copy is a closed embedding: a closed endpoint subset has saturation itself if it misses the basepoint, and its union with the collapsed track if it contains it. Applying this to the free copies of A and X recovers i as a closed embedding. All cylinder homotopies descend with their actual topology by F5.