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\title[A contractible four-manifold with uniformly positive scalar curvature]
{A contractible open four-manifold with a complete metric of uniformly positive scalar curvature}


\author{anonymous contributor to mathematics}
\address{}
\email{}
\date{September 21, 2026}

\subjclass[2020]{53C21, 57R65, 57N13, 20E06}
\keywords{positive scalar curvature, contractible four-manifold, topology at infinity, Baumslag--Solitar group}

\begin{document}
	
	\begin{abstract}
		We construct a smooth contractible open four-manifold that is not homeomorphic to $\R^4$
		and admits a complete Riemannian metric with scalar curvature at least one. This gives a negative answer to a question raised by
		Chang--Weinberger--Yu and Chodosh--M\'aximo--Mukherjee.
	\end{abstract}
	
	\maketitle
	
	\section{Introduction}\label{sec:introduction}
	
	A complete metric with uniformly positive scalar curvature imposes restrictions on the
	topology of an open manifold, including its topology at infinity. For contractible
	manifolds, these restrictions are particularly strong in dimension three:
	Chang--Weinberger--Yu proved that a contractible open three-manifold admitting such a
	metric is homeomorphic to $\R^3$~\cite[Theorem~1]{CWY10}. The standard $\R^n$, for $n\geq3$, admits a complete metric with uniformly positive scalar curvature, for instance the torpedo metric; for its construction, see \cite[Section~2.3]{EF21}.
	
	In dimension four, Chodosh--M\'aximo--Mukherjee proved that if the interior of a compact
	contractible smooth four-manifold admits a complete metric with uniformly positive
	scalar curvature, then the compact manifold is homeomorphic to the
	four-ball~\cite[Theorem~1.2]{CMM24}. For Mazur manifolds, they obtained a diffeomorphism
	conclusion~\cite[Theorem~1.1]{CMM24}. The question whether the standard $\R^4$ is the only
	contractible smooth four-manifold admitting a complete
	metric with uniformly positive scalar curvature was raised
	by Chang--Weinberger--Yu~\cite[Section~5]{CWY10} and
	reiterated by
	Chodosh--M\'aximo--Mukherjee~\cite[Question~1.5]{CMM24}.
	
	The following construction gives a negative answer.
	
	\begin{theorem}\label{thm:main}
		There exists a smooth, connected, one-ended, contractible open four-manifold $M$
		carrying a complete Riemannian metric $g$ with $\Scal_g\geq1$. The manifold $M$ is not
		simply connected at infinity and hence is not homeomorphic to $\R^4$.
	\end{theorem}
	
	We obtain a closed four-manifold $Z$ by circle surgery on $\#^2(S^1\times S^3)$, and
	then remove tubular neighborhoods of two circles to produce a compact cobordism $W$
	between two copies of $S^1\times S^2$. We join copies $W_0,W_1,\ldots$ in a sequence and
	cap the first boundary with $C=S^2\times D^2$; see Figure~\ref{fig:construction}.
	
	\begin{figure}[!htbp]
		\centering
		\begin{tikzpicture}[x=1cm,y=.95cm,font=\small,>=Stealth,line width=.55pt]
			% These are incidence diagrams, not pictures of an embedded four-manifold.
			\node at (1.05,2.65) {$\#^2(S^1\times S^3)$};
			\draw[->] (2.65,2.65) -- node[above,font=\scriptsize,align=center]
			{circle surgery} (4.3,2.65);
			\node at (4.7,2.65) {$Z$};
			\draw[->] (5.15,2.65) -- node[above,font=\scriptsize,align=center]
			{remove two tubular\\neighborhoods} (8.2,2.65);
			\node at (8.7,2.65) {$W$};
			\node[font=\scriptsize] at (8.7,2.18) {$\partial W=Y_t\sqcup Y_a$};
			
			% The cap and successive copies share boundary cross-sections.
			\draw (1.9,.62) -- (.62,.62)
			.. controls (.1,.62) and (.1,-.62) .. (.62,-.62) -- (1.9,-.62);
			\draw (1.9,.62) -- (9.4,.62);
			\draw (1.9,-.62) -- (9.4,-.62);
			\foreach \x in {1.9,4.4,6.9,9.4} {\draw (\x,-.62) -- (\x,.62);}
			\node[align=center] at (1.04,0) {$C$\\$S^2\times D^2$};
			\node at (3.15,0) {$W_0$};
			\node at (5.65,0) {$W_1$};
			\node at (8.15,0) {$W_2$};
			\draw[densely dashed] (9.4,.62) -- (10.3,.62);
			\draw[densely dashed] (9.4,-.62) -- (10.3,-.62);
			\draw[->] (9.65,0) -- (10.7,0);
			\node at (11,0) {$\cdots$};
			\node[above,font=\scriptsize] at (1.9,.73) {$\partial C\sim Y_t^0$};
			\node[above,font=\scriptsize] at (4.4,.73) {$Y_a^0\sim Y_t^1$};
			\node[above,font=\scriptsize] at (6.9,.73) {$Y_a^1\sim Y_t^2$};
			\node[above,font=\scriptsize] at (9.4,.73) {$Y_a^2\sim Y_t^3$};
			\draw[decorate,decoration={brace,mirror,amplitude=4pt}]
			(.3,-.88) -- node[below=4pt] {$K_0$} (1.9,-.88);
			\draw[decorate,decoration={brace,mirror,amplitude=4pt}]
			(.3,-1.62) -- node[below=4pt] {$K_1$} (4.4,-1.62);
			\draw[decorate,decoration={brace,mirror,amplitude=4pt}]
			(.3,-2.36) -- node[below=4pt] {$K_2$} (6.9,-2.36);
		\end{tikzpicture}
		\caption{The construction of $W$ and the infinite union. The incoming and outgoing
			boundaries of $W_j$ are denoted by $Y_t^j$ and $Y_a^j$. Each vertical seam represents an
			entire $S^1\times S^2$ boundary, and $\sim$ denotes identification by the gluing map.
			The compact set $K_n$ consists of the cap and the first $n$ copies of $W$. The outlines
			show only how the pieces are attached; they do not assert that $W$ is a product
			cobordism.}
		\label{fig:construction}
	\end{figure}
	
	Let $K_n$ be the compact manifold formed by the cap $C$ and the
	first $n$ copies of $W$. Each $K_n$ is simply connected, with
	$H_2(K_n;\Z)\cong\Z$ and all other reduced integral homology groups
	equal to zero. Attaching the next copy of $W$ makes every class
	in $H_2(K_n;\Z)$ null-homologous, while providing a new generator
	for $H_2(K_{n+1};\Z)$. Since every singular cycle in $M$ lies in
	a finite stage, all reduced integral homology groups of $M$
	vanish. The manifold $M$ is also simply connected, so the
	Hurewicz and Whitehead theorems imply that it is contractible.
	
	On the other hand, both boundary inclusions into $W$ induce
	injective homomorphisms on fundamental groups. Consequently,
	outside every compact subset of $M$ there is a loop that cannot
	be contracted in $M\setminus C$: although it contracts in $M$,
	every contraction must meet the cap. Thus $M$ is not simply
	connected at infinity and is not homeomorphic to $\R^4$.
	
	The metric is constructed on the same pieces. The Gromov--Lawson surgery construction
	and standardization near the two circles give $W$ a positive-scalar-curvature metric
	with identical product metrics near its boundary components. Copies of this metric fit
	together smoothly. Repeating a single compact metric gives a uniform positive lower
	bound for scalar curvature, and crossing each new copy requires a fixed positive amount
	of length. The resulting metric is complete.
	
	Our example does not arise as the interior of a compact smooth four-manifold. Indeed, suppose that $M$ were diffeomorphic to $\Int N$ for some compact smooth four-manifold $N$. Since the inclusion $\Int N\hookrightarrow N$ is a homotopy equivalence and $M$ is contractible, $N$ would also be contractible. The complete uniformly psc metric on $M$ would then satisfy the hypotheses of Chodosh--M\'aximo--Mukherjee~\cite[Theorem~1.2]{CMM24}, which imply that $N$ is homeomorphic to $D^4$. Hence $M$ would be homeomorphic to $\R^4$, contradicting Theorem~\ref{thm:main}. Thus the hypothesis that the open manifold is the interior of a compact manifold is essential in their result.
	
	Section~\ref{sec:topology} gives the topological construction, and
	Section~\ref{sec:metric} constructs the metric. All homology groups below have integral
	coefficients. A metric is called \emph{psc} if its scalar curvature is positive, and
	\emph{uniformly psc} if the scalar curvature has a positive constant lower bound.
	
	
	
	\section{The topological construction}\label{sec:topology}
	
	\subsection{The compact cobordism}\label{subsec:cobordism}
	
	The construction uses the Baumslag--Solitar group~\cite{BS62}
	\begin{align}\label{eq:group}
		G=BS(1,2)=\langle a,t\mid tat^{-1}=a^2\rangle.
	\end{align}
	Its abelianization is generated by $[t]$, whereas $[a]=0$. On the other hand, setting
	$t=1$ in the defining relation forces $a=1$, so $G/\normal{t}=1$. Both $a$ and $t$ have
	infinite order: the affine transformations $a(x)=x+1$ and $t(x)=2x$ satisfy the relation
	and have infinite order. These elementary properties will have different consequences
	for finite unions and for neighborhoods of infinity.
	
	Start with $X=\#^2(S^1\times S^3)$ and identify $\pi_1(X)$ with the free group $F(a,t)$. Choose pairwise disjoint smoothly embedded circles $c_a,c_t,c_r$ representing the conjugacy classes of $a,t$, and $r=tat^{-1}a^{-2}$, respectively. General position allows these representatives to be chosen embedded and pairwise disjoint. Their normal bundles are oriented rank-three bundles over $S^1$ and hence are trivial. Choose framings, and let $\nu(c_i)\cong S^1\times D^3$ denote the corresponding tubular neighborhoods. Perform circle surgery along $c_r$ by setting
	\begin{align}\label{eq:Z}
		Z=\bigl(X\setminus\Int\nu(c_r)\bigr)
		\cup_{S^1\times S^2}(D^2\times S^2).
	\end{align}
	Since $c_a$ and $c_t$ are disjoint from $\nu(c_r)$, after shrinking the tubular neighborhoods if necessary, they remain embedded circles in $Z$.
	
	\begin{lemma}\label{lem:Z}
		The closed oriented four-manifold $Z$ admits a psc metric, has fundamental group $G$,
		and has the integral homology of $S^1\times S^3$. The class $[c_t]$ generates $H_1(Z)$,
		and $[c_a]=0$.
	\end{lemma}
	
	\begin{proof}
		The product $S^1\times S^3$ admits a psc metric. The Gromov--Lawson connected-sum and
		surgery constructions therefore give psc metrics on $X$ and $Z$, since the surgery
		in~\eqref{eq:Z} has codimension three~\cite[Theorem~A]{GL80}.
		
		Removing an embedded circle from a four-manifold does not change the fundamental group:
		loops and homotopies of loops can be made disjoint from the circle by relative
		transversality, since their dimensions are at most two. Removing a tubular neighborhood
		has the same effect, by radial retraction. Van Kampen's theorem applied to~\eqref{eq:Z}
		thus gives $\pi_1(Z)=F(a,t)/\normal{r}=G$. The assertions about $H_1(Z)$ follow by
		abelianization.
		
		We have $\chi(X)=-2$, and circle surgery increases the Euler characteristic by two. Thus
		$\chi(Z)=0$. Poincar\'e duality gives $b_3(Z)=b_1(Z)=1$, and the Euler characteristic
		then gives $b_2(Z)=0$. Since $H_1(Z)\cong\Z$, the universal coefficient theorem
		identifies $H^2(Z)$ with $\operatorname{Hom}(H_2(Z),\Z)$. The latter vanishes because
		$H_2(Z)$ is finitely generated of rank zero. Duality now gives $H_2(Z)=0$, including its
		torsion. The remaining groups follow from connectedness and duality.
	\end{proof}
	
	Choose disjoint framed tubular neighborhoods
	$N_t,N_a\cong S^1\times D^3$ of $c_t,c_a$ in $Z$, and set
	\begin{align}\label{eq:W}
		W=Z\setminus\Int(N_t\sqcup N_a),\qquad
		\partial W=Y_t\sqcup Y_a.
	\end{align}
	The framings identify both boundary components with $S^1\times S^2$. We regard $Y_t$ as the incoming boundary and $Y_a$ as the outgoing boundary. In $Y_i$, write $\lambda_i=S^1\times{*}$ for the longitude and $\mu_i={*}\times S^2$ for the meridian sphere. If either tubular neighborhood is replaced by a smaller concentric one with respect to the chosen framing, the complement changes only by the addition of a product collar $(S^1\times S^2)\times[0,1]$. In particular, the resulting framed complement is diffeomorphic to $W$, and the boundary identifications, including the longitude and meridian classes, are unchanged up to isotopy. 
	
	
	\begin{lemma}\label{lem:W}
		The cobordism $W$ is connected, with $\pi_1(W)\cong G$. With suitable
		base paths, the boundary inclusions induce injective homomorphisms on
		fundamental groups, sending $\lambda_t,\lambda_a$ to $t,a$, respectively.
		
		The inclusion $Y_t\hookrightarrow W$ induces isomorphisms on $H_1$ and
		$H_3$ and the zero map on $H_2$, while $Y_a\hookrightarrow W$ induces
		an isomorphism on $H_2$.
	\end{lemma}
	
	\begin{proof}
		General position shows that $W$ is connected and that
		$W\hookrightarrow Z$ induces an isomorphism on fundamental groups.
		Under the identification $\pi_1(Z)=G$ from Lemma~\ref{lem:Z}, the
		longitudes represent $t,a$. Their infinite orders imply injectivity
		of the boundary homomorphisms. Abelianization gives the assertion
		about $H_1$.
		
		Excision and the Thom isomorphism give
		$H_k(Z,W)\cong H_{k-3}(c_t\sqcup c_a)$, so the only nonzero relative
		groups are $\Z^2$ in degrees four and three. Order their generators
		by $t,a$. After choosing orientations, the map
		$H_4(Z)\to H_4(Z,W)$ is $1\mapsto(1,1)$. The map
		$H_3(Z)\to H_3(Z,W)$ records intersection numbers with the two
		circles and is $1\mapsto(1,0)$ by Lemma~\ref{lem:Z}. Since
		$H_2(Z)=0$ and both maps are injective, the long exact sequence
		gives $H_4(W)=0$ and
		\begin{align*}
			H_3(W)&\cong\Z^2/\langle(1,1)\rangle,\\
			H_2(W)&\cong\Z^2/\langle(1,0)\rangle.
		\end{align*}
		Under the connecting homomorphisms, the standard basis vectors map,
		up to signs, to the images of $[Y_t],[Y_a]$ in degree three and of
		$[\mu_t],[\mu_a]$ in degree two. Thus the image of $[Y_t]$ generates
		$H_3(W)$, while $[\mu_t]$ maps to zero and $[\mu_a]$ maps to a
		generator of $H_2(W)$. This proves the remaining assertions.
	\end{proof}
	
	\subsection{The infinite union}\label{subsec:union}
	
	Let $W_0,W_1,\ldots$ be copies of $W$, with boundary components $Y_t^j,Y_a^j$. Glue
	$Y_a^j$ to $Y_t^{j+1}$ by the product diffeomorphism
	\begin{align}\label{eq:phi}
		\phi(e^{i\theta},x)=(e^{-i\theta},x).
	\end{align}
	Choose the boundary product coordinates consistently so that $\phi$ reverses the induced
	boundary orientations. Cap $Y_t^0$ by $C=S^2\times D^2$, using the product map
	$\psi(x,e^{i\theta})=(e^{-i\theta},x)$ from $\partial C$ to $Y_t^0$. Orient $C$ so that
	$\psi$ reverses the boundary orientations as well.
	
	Set $K_0=C$ and define
	\begin{align}\label{eq:exhaustion}
		K_n=C\cup\bigcup_{j=0}^{n-1}W_j\quad(n\geq1),
		\qquad M=\bigcup_{n\geq0}K_n.
	\end{align}
	All gluings use collars. Each seam consequently has a neighborhood diffeomorphic to
	$(S^1\times S^2)\times(-\varepsilon,\varepsilon)$. The pieces form a locally finite
	chain, and there is no point added at its infinite end. Thus $M$ is a smooth open
	four-manifold, $K_n\subset\Int K_{n+1}$, and every compact subset of $M$ lies in the
	interior of some $K_n$. For $n\geq1$, the only boundary component of $K_n$ is
	$Y_a^{n-1}$.
	
	\begin{lemma}\label{lem:finite-stages}
		Each $K_n$ is simply connected, has the homology of $S^2\times D^2$, and its boundary
		inclusion induces an isomorphism on $H_2$. The inclusion $K_n\hookrightarrow K_{n+1}$
		induces the zero map on $H_2$.
	\end{lemma}
	
	\begin{proof}
		The first three assertions hold for $K_0=C$. Assume they hold for $K_n$ and consider
		$K_{n+1}=K_n\cup_{Y_t^n}W_n$. The incoming longitude represents $t_n$ in $\pi_1(W_n)$
		and is null-homotopic in $K_n$. Van Kampen's theorem gives
		\begin{align*}
			\pi_1(K_{n+1})\cong G/\normal{t}=1.
		\end{align*}
		
		For homology, use the Mayer--Vietoris sequence with collar neighborhoods of the two
		pieces. The maps from the common boundary $Y\cong S^1\times S^2$ to their direct-sum
		homology groups are, up to signs,
		\begin{align*}
			H_3(Y)&\longrightarrow H_3(K_n)\oplus H_3(W_n),& x&\longmapsto(0,x),\\
			H_2(Y)&\longrightarrow H_2(K_n)\oplus H_2(W_n),& x&\longmapsto(x,0),\\
			H_1(Y)&\longrightarrow H_1(K_n)\oplus H_1(W_n),& x&\longmapsto(0,x).
		\end{align*}
		The degree-three map is an isomorphism and the degree-two map is injective. Exactness
		therefore gives $H_4(K_{n+1})=H_3(K_{n+1})=0$. The degree-one map is also an isomorphism. Hence exactness gives
		\begin{align*}
			H_2(K_{n+1})
			&\cong
			\frac{H_2(K_n)\oplus H_2(W_n)}
			{H_2(K_n)\oplus\{0\}}
			\cong H_2(W_n).
		\end{align*}
		Under this identification, the inclusion $W_n\hookrightarrow K_{n+1}$ induces an
		isomorphism on $H_2$, while the inclusion $K_n\hookrightarrow K_{n+1}$ induces the zero
		map. Connectedness and simple connectivity give the assertions in degrees zero and one,
		and the remaining homology groups vanish.
		
		Finally, $\partial K_{n+1}=Y_a^n$, and the boundary inclusion
		factors as
		$Y_a^n\hookrightarrow W_n\hookrightarrow K_{n+1}$.
		The first inclusion induces an isomorphism on $H_2$ by
		Lemma~\ref{lem:W}, and the second does so by the preceding
		calculation. Thus the boundary inclusion also induces an
		isomorphism on $H_2$. This completes the induction and proves
		the last assertion for every $n\geq0$.
	\end{proof}
	
	\begin{proposition}\label{prop:contractible}
		The manifold $M$ is contractible and has exactly one end.
	\end{proposition}
	
	\begin{proof}
		Every loop in $M$ has compact image and lies in a finite stage, where it contracts by
		Lemma~\ref{lem:finite-stages}. Thus $\pi_1(M)=1$. Singular chains also have compact
		support, so homology commutes with this increasing union. In degree two,
		Lemma~\ref{lem:finite-stages} gives
		\begin{align*}
			H_2(M)\cong\varinjlim\bigl(\Z\xrightarrow{0}\Z\xrightarrow{0}\Z\xrightarrow{0}\cdots\bigr)=0.
		\end{align*}
		All other reduced homology groups vanish as well. A simply connected acyclic space of CW
		homotopy type is contractible, by the Hurewicz and Whitehead theorems. These apply
		because $M$ is a smooth manifold.
		
		For every $n$, the complement of $\Int K_n$ is the connected chain
		$W_n\cup W_{n+1}\cup\cdots$. Its interior in $M$ is connected as well.
		Moreover, every neighborhood of infinity contains one of these complements,
		so $M$ has one end.
	\end{proof}
	
	\subsection{The topology at infinity}\label{subsec:infinity}
	
	A manifold is \emph{simply connected at infinity} if for every compact subset $A$ there
	is a compact subset $B\supset A$ such that every loop outside $B$ contracts outside $A$.
	We show that the manifold $M$ fails this condition.
	
	Set $U_n=M\setminus\Int K_n$ and $U_{n,m}=W_n\cup\cdots\cup W_m$ for $m\geq n$. Choose
	compatible base paths along a proper ray through the pieces, and write $G_j=\pi_1(W_j)$.
	The gluing map~\eqref{eq:phi} identifies $a_j$ with $t_{j+1}^{-1}$. Thus van Kampen's
	theorem gives
	\begin{align}\label{eq:amalgam}
		\pi_1(U_{n,m})\cong
		G_n*_{\langle a_n\rangle=\langle t_{n+1}^{-1}\rangle}G_{n+1}
		*\cdots*_{\langle a_{m-1}\rangle=\langle t_m^{-1}\rangle}G_m
	\end{align}
	when $m>n$, and $\pi_1(U_{n,n})=G_n$.
	
	\begin{proposition}\label{prop:infinity}
		The manifold $M$ is not simply connected at infinity and is not
		homeomorphic to $\R^4$.
	\end{proposition}
	
	\begin{proof}
		Fix $n\geq0$. By Lemma~\ref{lem:W}, both boundary inclusions into $W$
		induce injections on fundamental groups. The normal form theorem
		for free products with amalgamation therefore shows that each factor
		$G_j$, $n\leq j\leq m$, in~\eqref{eq:amalgam} embeds in
		$\pi_1(U_{n,m})$, and that each inclusion
		$U_{n,m}\hookrightarrow U_{n,m+1}$ induces an injection on
		fundamental groups~\cite[I, \S1.2, Theorem~1]{SerreTrees}.
		By local finiteness, the compact image of any loop or null-homotopy
		of a loop in $U_n$ is contained in some $U_{n,m}$. Hence the inclusion
		$W_j\hookrightarrow U_n$ induces an injection
		$G_j\hookrightarrow\pi_1(U_n)$ for every $j\geq n$.
		
		Let $B\supset K_n$ be any compact subset of $M$, and choose $m>n$
		such that $B\subset\Int K_m$. Push a longitude of $Y_a^m$
		representing $a_m$ into $\Int W_m$, and denote the resulting loop
		by $\gamma_m$. Since $\Int W_m\cap K_m=\varnothing$, this loop
		lies outside $B$. The element $a_m$ is nontrivial in $G_m$, so
		$\gamma_m$ is not null-homotopic in $U_n$. Since
		$M\setminus K_n\subset U_n$, it is not null-homotopic in
		$M\setminus K_n$ either. As $B$ was arbitrary, this disproves
		simple connectivity at infinity with the fixed compact set $A=K_n$.
		
		The complement of every closed Euclidean ball in $\R^4$ is simply
		connected, so $\R^4$ is simply connected at infinity. Since this
		property is preserved by homeomorphisms, $M$ is not homeomorphic
		to $\R^4$.
	\end{proof}
	
	\section{The metric construction}\label{sec:metric}
	
	It remains to put a complete uniformly psc metric on $M$. We first construct a psc metric on $W$ with the same product metric near both boundary components. We can then put an isometric copy of this metric on every $W_j$ and glue the copies smoothly along their product collars. In particular, the scalar curvature lower bound and the distance between the two boundary components of each $W_j$ are independent of $j$.
	
	\begin{lemma}\label{lem:metric-W}
		The cobordism $W$ admits a psc metric $g_W$ with the same product
		metric near both boundary components. More precisely, there are
		constants $\varepsilon,\ell,\rho>0$ such that each $Y_i$, $i=t,a$,
		has a product neighborhood $[0,\varepsilon)\times Y_i$ in $W$ on
		which
		\begin{align}\label{eq:collar}
			g_W=ds^2+g_Y,
			\qquad
			g_Y=\ell^2d\theta^2+\rho^2g_{S^2}.
		\end{align}
		Here $s=0$ is the boundary and $s$ increases into $W$. The formula
		for $g_Y$ uses the fixed identification of each $Y_i$ with
		$S^1\times S^2$ from the topological construction. The variable
		$\theta\in\R/2\pi\Z$ is the circle coordinate, and $g_{S^2}$ is the
		unit round metric on $S^2$.
	\end{lemma}
	
	\begin{proof}
		By Lemma~\ref{lem:Z}, the closed manifold $Z$ admits a psc metric.
		We will arrange that this metric is a product on the outside of each
		deleted tubular neighborhood.
		
		The fixed tubular coordinates identify $N_t$ and $N_a$ with copies
		of $S^1\times D^3$, where $D^3$ is the closed unit ball. Extend these
		coordinates slightly beyond the boundaries of $N_t$ and $N_a$,
		keeping the two enlarged neighborhoods disjoint. A radial
		reparametrization outside $D^3$ then gives an open embedding
		\begin{align*}
			\Phi:(S^1\sqcup S^1)\times\R^3\longrightarrow Z
		\end{align*}
		that agrees with the fixed parametrizations on
		$(S^1\sqcup S^1)\times D^3$. Thus $r=1$ corresponds to
		$Y_t\sqcup Y_a$, where $r$ is the radial coordinate in $\R^3$.
		
		Choose $\ell>0$ and a standard torpedo metric on $\R^3$ of the form
		\begin{align*}
			g_{\mathrm{tor}}
			=dr^2+\eta(r)^2g_{S^2},
		\end{align*}
		with positive scalar curvature and $\eta(r)=\rho$ for $r\geq r_0$,
		where $0<r_0<1$. Such a choice is obtained by scaling a standard
		torpedo metric. Use the same metric $\ell^2d\theta^2$ on both circles
		and the same $g_{\mathrm{tor}}$ on both normal factors. Since the
		circle metric is flat, the product
		$\ell^2d\theta^2+g_{\mathrm{tor}}$ has positive scalar curvature.
		
		Fix $R>1$. Applying the standardization theorem to the two circles
		together, which have codimension three in $Z$, gives a psc metric
		$g_Z$ such that
		\begin{align}\label{eq:standardization}
			\Phi^*g_Z
			=
			\ell^2d\theta^2+g_{\mathrm{tor}}
			\quad\text{on }(S^1\sqcup S^1)\times B_R^3;
		\end{align}
		see \cite[Theorem~1.2 and Remark~1.3]{EF21}. Here $B_R^3$ is the
		open ball of radius $R$.
		
		Set $g_W=g_Z|_W$. In these coordinates, the regions $1\leq r<R$
		lie in $W$. Since $r_0<1$, their metric is exactly
		\begin{align*}
			dr^2+\ell^2d\theta^2+\rho^2g_{S^2}.
		\end{align*}
		Taking $s=r-1$ and $\varepsilon=R-1$ gives the required product
		neighborhoods. At $s=0$, the circle and sphere coordinates are the
		fixed boundary coordinates. Both neighborhoods use the same
		constants $\ell$ and $\rho$.
	\end{proof}
	
	We next equip the cap $C=S^2\times D^2$ with a psc metric whose
	product collar matches that of $W$. Choose a smooth nonincreasing
	function $q:[0,1]\to[0,1]$ which equals $1$ near zero and $0$ near one,
	and set
	\begin{align*}
		L
		=
		\frac{\ell}{\displaystyle\int_0^1 q(u)\,du},
		\quad
		\zeta(r)
		=
		\int_0^r q(t/L)\,dt,
		\qquad 0\leq r\leq L.
	\end{align*}
	Then $\zeta(r)=r$ near zero, $\zeta(r)=\ell$ near $L$, and
	$\zeta''\leq0$.
	
	In polar coordinates on the disk, with boundary $r=L$ and angular
	coordinate $\theta$, define
	\begin{align*}
		h_D=dr^2+\zeta(r)^2d\theta^2.
	\end{align*}
	Since $\zeta(r)=r$ near zero, the metric $h_D$ is Euclidean near the
	center and extends smoothly across it. For $r>0$, its Gaussian curvature
	is
	\begin{align*}
		K_{h_D}=-\frac{\zeta''}{\zeta}\geq0.
	\end{align*}
	Hence the product metric $g_C=\rho^2g_{S^2}+h_D$ has scalar curvature
	\begin{align*}
		\Scal_{g_C}
		=
		\frac{2}{\rho^2}
		-\frac{2\zeta''}{\zeta}
		\geq
		\frac{2}{\rho^2}
		>0
	\end{align*}
	for $r>0$. Near the disk center, $h_D$ is flat, so
	$\Scal_{g_C}=2/\rho^2$ there as well. Near $\partial C$, setting
	$u=L-r$ gives
	\begin{align*}
		g_C
		=
		du^2+\rho^2g_{S^2}+\ell^2d\theta^2.
	\end{align*}
	
	The maps $\phi:Y_a^j\to Y_t^{j+1}$ used to glue consecutive
	copies of $W$, and $\psi:\partial C\to Y_t^0$ used to attach the cap,
	are isometries for these boundary metrics. Put a copy of $g_W$ on every
	$W_j$ and put $g_C$ on $C$, and glue the pieces along these maps using
	the product collars. Near each
	gluing hypersurface, use the negative of the inward collar coordinate
	on one side and the positive inward collar coordinate on the other.
	After identifying the boundary coordinates by the gluing map, the metric
	on both sides is
	\begin{align*}
		ds^2+g_Y.
	\end{align*}
	It is therefore smooth across each gluing hypersurface and has positive
	scalar curvature there. Since the pieces form a locally finite chain,
	this defines a smooth psc metric $g$ on $M$.
	
	\begin{proof}[Proof of Theorem~\ref{thm:main}]
		Propositions~\ref{prop:contractible} and~\ref{prop:infinity}
		give the topological assertions. Since $g_C$ and $g_W$
		have positive scalar curvature on the compact manifolds
		$C$ and $W$, respectively, the constant
		\begin{align*}
			\sigma
			=\min\left\{
			\min_C\Scal_{g_C},\,
			\min_W\Scal_{g_W}
			\right\}
		\end{align*}
		is positive. Every $W_j$ carries an isometric copy of $g_W$,
		so the glued metric satisfies $\Scal_g\geq\sigma$ on $M$.
		
		To prove completeness, fix $p\in C$ and let
		$\delta=\operatorname{dist}_{g_W}(Y_t,Y_a)>0$, where the
		distance is intrinsic to $W$. Fix $N\geq1$, let
		$q\in M\setminus K_N$, and let $\gamma$ be any piecewise
		smooth curve from $p$ to $q$. For each $j=0,\ldots,N-1$,
		consider the first time that $\gamma$ reaches $Y_a^j$,
		and take the segment from its last preceding visit to
		$Y_t^j$ up to that time. Since $\gamma$ can leave $W_j$
		only through $Y_t^j$ or $Y_a^j$, this segment lies in
		$W_j$ and has length at least $\delta$. These segments
		occur in order and have disjoint parameter interiors,
		so $L_g(\gamma)\geq N\delta$. Taking the infimum over
		all such curves gives $d_g(p,q)\geq N\delta$.
		
		For every $R>0$, choose $N$ so that $N\delta>R$. Then
		$\overline B_g(p,R)\subset K_N$. Since $K_N$ is compact
		and the ball is closed in $K_N$, the ball is compact.
		Thus $(M,g)$ is proper, and the Hopf--Rinow theorem
		gives completeness.
		
		Finally, set $\widetilde g=\sigma g$. Constant rescaling
		preserves completeness and gives
		$\Scal_{\widetilde g}=\sigma^{-1}\Scal_g\geq1$.
	\end{proof}
	
	\medskip
	\noindent\textbf{On the use of AI.}
	In a series of conversations concerning the above question, GPT-6 constructed a highly intricate counterexample and developed the associated proofs. Its contributions include the topological construction and the metric construction. The human researchers’ work consists of posing the research problem, simplifying the counterexample construction, completing the writing of this manuscript, and interpreting and verifying GPT-6’s arguments. The human authors take full responsibility for the content, accuracy, and integrity of this manuscript. The authors will not submit this manuscript for publication.
	
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\end{document}