Compact topological manifold boundaries admit collars

The interior is open by [F1]: in a chart an interior point has a smaller Euclidean ball missing the model boundary. Thus $A$ is closed and compact. Restricting each boundary chart to the model hyperplane gives charts for the Hausdorff second-countable boundary $A$, of dimension $n-1$ and without boundary. We can take the ambient boundary charts $W_i$ homeomorphic to all of ${\mathbb{R}}^{n-1}\times [0,\mathrm{\infty })$. Indeed a product $B_r^{n-1}\times [0,r)$ inside an original chart can be sent there by the horizontal map $y\mapsto y\mathrm{/}(r-\parallel y\parallel )$ and vertical map $v\mapsto v\mathrm{/}(r-v)$; their inverses are $z\mapsto rz\mathrm{/}(1+\parallel z\parallel )$ and $t\mapsto rt\mathrm{/}(1+t)$.

Attach an external collar by defining $M^{\mathrm{\prime }}=(M\bigsqcup (A\times [0,1]))\mathrm{/}((x,0)\sim x)$. The original $M$ and the external collar embed with their usual topologies. For example an open set $O\subset M$ extends to the quotient-open set consisting of $O$ and $(O\cap A)\times [0,ϵ)$ for $0<ϵ\le 1$; positive external heights have their ordinary product neighborhoods. These descriptions also prove that $M^{\mathrm{\prime }}$ is Hausdorff. Two points in $M$ can use disjoint open neighborhoods extended in this way; two external points have separated base or height neighborhoods; to separate an external point of height $t>0$ from a point in $M$, use a collar cutoff $ϵ<t$ for the latter and an external neighborhood above $ϵ$ for the former. Compatibility at height zero is exactly the quotient topology in [F4]. Each half-space chart from step 1.1, joined to its external part, therefore has signed coordinates $(x,s)$ with $s\le 0$ inside $M$ and $0\le s\le 1$ outside, the two copies at zero identified. The map across zero and its inverse are continuous by finite closed pasting and [F4].

There are finitely many continuous functions $b_i:A\to [0,1]$ whose positive sets cover $A$ and whose supports are compact subsets of $A\cap W_i$. To construct them, in any boundary chart take the radial function $\max (0,1-\parallel y-y_0\parallel \mathrm{/}r)$ with its closed radius-$r$ ball contained in that chart, and extend it by zero on the rest of $A$. Its support is compact by [F2], hence closed, and is contained in the chart; at a point off that compact support an open neighborhood has function identically zero, proving continuity of the extension. The positive sets of all such functions cover $A$, so compactness gives a finite subcover and hence finitely many functions. For $n=1$, boundary charts are singletons and use the constant value-one function there, extended by zero. Set ${\phi }_i=b_i\mathrm{/}({\sum }_j{b}_j)$. The denominator is positive everywhere, so these are continuous, have the same compact supports and sum to $1$.

Put ${\psi }_k={\phi }_1+\cdots +{\phi }_k$, ${\psi }_0=0$, and let $M_k\subset M^{\mathrm{\prime }}$ consist of $M$ and the external points $(x,t)$ with $0\le t\le {\psi }_k(x)$. Thus $M_0=M$ and $M_m=M^{\mathrm{\prime }}$ for the finite number $m$ of functions. In the signed coordinates for $W_k$ take its inner collar with $-1\le s\le 0$. Define $h_k:M_{k-1}\to M_k$ to be identity outside this collar and the external segments over $A\cap W_k$, and on their part with $-1\le s\le {\psi }_{k-1}(x)$ set $h_k(x,s)=(x,-1+(s+1)\frac{1+{\psi }_k(x)}{1+{\psi }_{k-1}(x)})\mathrm{.}$ The denominator is at least one. This increasing affine map sends the bottom $-1$ to $-1$ and the old top ${\psi }_{k-1}(x)$ to the new top ${\psi }_k(x)$. If ${\phi }_k(x)=0$ it is the identity on the entire segment.

The formula in step 3.1 defines a homeomorphism globally. In its signed-coordinate region it is continuous, including across zero; along the bottom it agrees with the identity on the region below. All points where it can differ from identity lie over the compact base support of ${\phi }_k$. The product of that support with $[-1,1]$ is compact by [F3], and its embedded image in the Hausdorff $M^{\mathrm{\prime }}$ is closed by [F2]. Away from this set the map is identity; at a point of the set all local pasting is within the signed-coordinate chart. Thus there is no continuity issue at the edge of the chart. The inverse has the same support and the formula $h_k^{-1}(x,s^{\mathrm{\prime }})=(x,-1+(s^{\mathrm{\prime }}+1)\frac{1+{\psi }_{k-1}(x)}{1+{\psi }_k(x)}),$ again continuous and fixing the bottom. Direct substitution proves both composites identities, and the increasing interval bijections plus the fixed complement prove bijectivity on the indicated $M_k$.

The finite composite $h=h_m\cdots h_1:M\to M^{\mathrm{\prime }}$ is a homeomorphism. For $x\in A$, the successive maps take the graph point $(x,{\psi }_{k-1}(x))$ to $(x,{\psi }_k(x))$, even when that stage's function is zero. Hence $h(x)=(x,1)$. The external region $A\times (0,1]$ is open in $M^{\mathrm{\prime }}$: its preimage under the quotient has empty intersection with $M$ and is open in $A\times [0,1]$. Therefore $c(x,t)=h^{-1}(x,1-t)(0\le t<1)$ is the required homeomorphism onto its open image $V$, and $c(x,0)=x$.

Fix $b=1\mathrm{/}2$. Within $c(A\times [0,b])$ define a homotopy by $c(x,v)\mapsto c(x,v+s(b-v)\mathrm{/}2)$ for $s\in [0,1]$, and use identity outside $c(A\times [0,b))$. The formulas agree at $v=b$ and are continuous by finite closed pasting: the first collar segment is compact by [F3], hence closed by [F2], and the complementary piece is closed because $c(A\times [0,b))$ is open. At $s=0$ this is identity, and at $s=1$ it sends every boundary point to positive height while preserving the interior throughout. Its terminal map $r:M\to M\setminus A$ is continuous into that subspace. The homotopy gives $ir\simeq 1_M$, and its restriction to the interior gives $ri\simeq 1_{M\setminus A}$, where $i$ is the inclusion. Thus $i$ is a homotopy equivalence.

For HEP put $b=1\mathrm{/}2$ again and retract $M\times I$ onto $T=M\times \{0\}\cup A\times I$. Outside $c(A\times [0,b))$ send $(y,t)$ to $(y,0)$. At $y=c(x,v)$, $0\le v<b$, put $a(v)=v\mathrm{/}(b-v)$. If $t\le a(v)$, send it to $(c(x,\frac{b(a(v)-t)}{1+a(v)-t}),0);$ if $t\ge a(v)$, send it to $(x,t-a(v))\in A\times I$. At equality both give $(x,0)$. At $v=0$ it is $(x,t)$, and at $t=0$ its first branch gives $(c(x,v),0)$. As $v$ approaches $b$ from below, $a(v)$ tends to infinity, uniformly dominating $t\in [0,1]$, and the new collar height tends to $b$; hence the formula extends continuously to the outside rule. The collar extends beyond $b$, so this limiting continuity is checked within its actual coordinate chart. These local formulas define a continuous map into the subspace $T$, fixing $T$ pointwise. To verify the unbased cofibration assertion directly, let $Z$ be any topological space, let $f:M\to Z$ and $g:A\times I\to Z$ be continuous, and suppose $g(x,0)=f(x)$ for every $x\in A$. Paste $f$ on $M\times \{0\}$ and $g$ on $A\times I$, the two closed pieces of $T$, and compose with the retraction. Finite closed pasting proves a continuous $H:M\times I\to Z$ satisfying $H(y,0)=f(y)$ and $H(x,t)=g(x,t)$ for all $x\in A$. This is the unbased homotopy extension property for every target $Z$, without appealing to a later definition.

Steps 5.1, 6.1 and 6.2 prove all assertions. Empty boundary, empty $M$ and dimension zero were settled in the Given paragraph; a compact zero-manifold has no boundary. One half-chart and zero individual bump functions cause no difficulty, since denominators in step 3.1 remain at least one and a zero increment is identity. The geometric endpoints $s=-1$, external height $1$, collar height $b$, and HEP times $0,1$ were checked explicitly. No chain or simplex convention is involved in this topological theorem. All covers are the families of every available chart bump, thinned only once to a finite subcover; all subsequent choices are finite and all maps have displayed formulas. No AC is used.