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I sketch the proof of Buoncristiano and Hacon:

Let $M$ be a parallelizable manifold of dimension $m$. Let $N$ be $M \times M \setminus U$, where $U$ is a tubular neighbourhood of the diagonal (invariant under the natural involution on $M\times M$.) The involution on $N$ can be induced from the antipodal involution on the sphere $S^q$ for a sufficiently big $q$ (i.e., one may chose a $\mathbb Z/2$-equivariant embedding $N\hookrightarrow S^q$). The boundary of $N$ is $M \times S^{m-1}$, and the involution on the boundary can be induced from that on $S^{m-1}$.

So factorizing out by the involution the manifold $N$ we get a manifold $N'$, its map $f$ to $RP^q$, and the boundary of $N'$ is mapped into $RP^{m-1} \subset RP^q$. Take an $RP^{q-m+1}$ in $RP^q$ that intersects $RP^{m-1}$ in a single point. If both $f$ and its restriction to the boundary are transverse to this $RP^{m-1}$ (what can be supposed) then $f^{-1}(RP^{q-m+1})$ is a manifold with boundary $M$. Q.E.D.

I sketch the proof of Buoncristiano and Hacon:

Let $M$ be a parallelizable manifold of dimension $m$. Let $N$ be $M \times M \setminus U$, where $U$ is a tubular neighbourhood of the diagonal (invariant under the natural involution on $M\times M$.) The involution on $N$ can be induced from the antipodal involution on the sphere $S^q$ for a sufficiently big $q$ (i.e., one may chose a $\mathbb Z/2$-equivariant embedding $N\hookrightarrow S^q$). The boundary of $N$ is $M \times S^{m-1}$, and the involution on the boundary can be induced from that on $S^{m-1}$.

So factorizing out by the involution the manifold $N$ we get a manifold $N'$, its map $f$ to $\mathbb{RP}^q$, and the boundary of $N'$ is mapped into $\mathbb{RP}^{m-1} \subset \mathbb{RP}^q$. Take an $\mathbb{RP}^{q-m+1}$ in $\mathbb{RP}^q$ that intersects $\mathbb{RP}^{m-1}$ in a single point. If both $f$ and its restriction to the boundary are transverse to this $\mathbb{RP}^{m-1}$ (this can be supposed) then $f^{-1}(\mathbb{RP}^{q-m+1})$ is a manifold with boundary $M$. Q.E.D.

I sketch the proof of Buoncristiano and Hacon:

Let $M$ be a parallelizable manifold of dimension $m$. Let $N$ be $M \times M \setminus U$, where $U$ is a tubular neighbourhood of the diagonal (invariant under the natural involution on $M\times M$.) The involution on $N$ can be induced from the antipodal involution on the sphere $S^q$ for a sufficiently big q. Since the boundary of $N$ is $M \times S^{m-1}$, hence the involution on the boundary can be induced from that on $S^{m-1}$.

So factorizing out by the involution the manifold $N$ we get a manifold $N'$, its map $f$ to $RP^q$, and the boundary of $N'$ is mapped into $RP^{m-1} \subset RP^q$. Take an $RP^{q-m+1}$ in $RP^q$ that intersects $RP^{m-1}$ in a single point. If both $f$ and its restriction to the boundary are transverse to this $RP^{m-1}$ (what can be supposed) then $f^{-1}(RP^{q-m+1})$ is a manifold with boundary $M$. Q.E.D.

I sketch the proof of Buoncristiano and Hacon:

Let $M$ be a parallelizable manifold of dimension $m$. Let $N$ be $M \times M \setminus U$, where $U$ is a tubular neighbourhood of the diagonal (invariant under the natural involution on $M\times M$.) The involution on $N$ can be induced from the antipodal involution on the sphere $S^q$ for a sufficiently big $q$ (i.e., one may chose a $\mathbb Z/2$-equivariant embedding $N\hookrightarrow S^q$). The boundary of $N$ is $M \times S^{m-1}$, and the involution on the boundary can be induced from that on $S^{m-1}$.

So factorizing out by the involution the manifold $N$ we get a manifold $N'$, its map $f$ to $RP^q$, and the boundary of $N'$ is mapped into $RP^{m-1} \subset RP^q$. Take an $RP^{q-m+1}$ in $RP^q$ that intersects $RP^{m-1}$ in a single point. If both $f$ and its restriction to the boundary are transverse to this $RP^{m-1}$ (what can be supposed) then $f^{-1}(RP^{q-m+1})$ is a manifold with boundary $M$. Q.E.D.

I sketch the proof of Buoncristiano and Hacon:

Let $M$ be a parallelizable manifold of dimension $m$. Let $N$ be $M \times M \setminus U$, where $U$ is a tubular neighbourhood of the diagonal (invariant under the natural involution on $M\times M$.) The involution on $N$ can be induced from the antipodal involution on the sphere $S^q$ for a sufficiently big . Since the boundary of $N$ is $M \times S^{m-1}$, hence the involution on the boundary can be induced from that on $S^{m-1}$.

So factorizing out by the involution the manifold $N$ we get a manifold $N'$, its map $f$ to $RP^q$, and the boundary of $N'$ is mapped into $RP^{m-1} \subset RP^q$. Take an $RP^{q-m+1}$ in $RP^q$ that intersects $RP^{m-1}$ in a single point. If both $f$ and its restriction to the boundary are transverse to this $RP^{m-1}$ (what can be supposed) then $f^{-1}(RP^{q-m+1})$ is a manifold with boundary $M$. Q.E.D.

I sketch the proof of Buoncristiano and Hacon:

Let $M$ be a parallelizable manifold of dimension $m$. Let $N$ be $M \times M \setminus U$, where $U$ is a tubular neighbourhood of the diagonal (invariant under the natural involution on $M\times M$.) The involution on $N$ can be induced from the antipodal involution on the sphere $S^q$ for a sufficiently big q. Since the boundary of $N$ is $M \times S^{m-1}$, hence the involution on the boundary can be induced from that on $S^{m-1}$.

So factorizing out by the involution the manifold $N$ we get a manifold $N'$, its map $f$ to $RP^q$, and the boundary of $N'$ is mapped into $RP^{m-1} \subset RP^q$. Take an $RP^{q-m+1}$ in $RP^q$ that intersects $RP^{m-1}$ in a single point. If both $f$ and its restriction to the boundary are transverse to this $RP^{m-1}$ (what can be supposed) then $f^{-1}(RP^{q-m+1})$ is a manifold with boundary $M$. Q.E.D.