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Let $1-Cob$ denote the category of oriented 0-manifolds and oriented cobordisms between them. If $W:A\to B$ is a cobordism, i.e. $\partial W\cong A+B$, we write $i^W_{dom}:A\to W$ and $i^W_{cod}:B\to W$ to denote the boundary-component inclusions. Let $\pi_0:{\bf Man}\to{\bf Set}$ denote the connected components functor.

Let $\mathcal{L}\subset Mor_{1-Cob}$ denote the set of morphisms $W$ for which the function $\pi_0(i^W_{cod}\,):\pi_0(B)\to \pi_0(W)$ is an injection and for which the 1-manifold $W$ contains no closed loops, $\not\exists (S^1\hookrightarrow W)$.

Let $\mathcal{R}\subset Mor_{1-Cob}$ denote the set of morphisms $W$ for which the function $\pi_0(i^W_{dom}):\pi_0(A)\to \pi_0(W)$ is an injection.

I believe that $\mathcal{L}$ and $\mathcal{R}$ form an orthogonal factorization system on $1-Cob$, but I don't know how to prove it.

Question: Can you find a proof, a reference, or a counterexample for this conjecture?




(Just for symmetry....)

Let $\mathcal{L'}\subset Mor_{1-Cob}$ denote the set of morphisms $W$ for which $\pi_0(i^W_{cod})$ is an injection.

Let $\mathcal{R'}\subset Mor_{1-Cob}$ denote the set of morphisms $W$ for which $\pi_0(i^W_{dom})$ is an injection and for which the 1-manifold $W$ contains no closed loops, $\not\exists (S^1\hookrightarrow W)$.

I also believe that $\mathcal{L'}$ and $\mathcal{R'}$ form an orthogonal factorization system on $1-Cob$, but I don't know how to prove it either.


Edit provenance: In an earlier version of this question, I had reversed $\mathcal{L}$ and $\mathcal{R}$. This allowed Chris Schommer-Pries to correctly answer my question as originally posed, but not the question I meant to ask (i.e., the one you see above now).

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I think you are right. The right class of your factorization system seems familiar. I believe it is the categorical model of the sphere spectrum introduced by Quillen: an object is a (finite) signed set and a morphism $f:X\to Y$ is a sign preserving injection together with an involution matching positive with negative elements in the complement of the image of $f$. If we neglect the loops, a cobordism $X\to Y$ is a span $X\leftarrow Z\rightarrow Y$ defined by two maps $s:Z\to X$ and $t:Z\to Y$ in this category. The loops can be pushed on either sides.

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    $\begingroup$ If this is André Joyal, then welcome to MathOverflow! $\endgroup$ – Todd Trimble Aug 5 '14 at 4:46
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    $\begingroup$ @ToddTrimble: If not, he is welcome as well :-) $\endgroup$ – Michal R. Przybylek Aug 6 '14 at 19:23
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[Note: This answer pertains to the orignal version of the question, and no longer applies given the subsequent edits].

To be a weak factorization system you need to factor morphisms as composites $f = r \circ \ell$, and you need there to exist solutions to certain lifting squares. I haven't thought about the existence of the necessary lifts, but the existence of a factorization follows by choosing a self-indexing Morse function (and dealing with the circle components separately). So it is possible your classes of morphisms form a weak factorization system.

However you asked about an orthogonal factorization system. For that you need that the solutions to the lifting problem are unique. This would also imply that the factorizations are unique. You can already see that that isn't the case. For example the circle can be built by gluing a cup (in L) to a cap (in R) or you can just build it all at once with a circle (in L) and an empty bordism (in R).

But we can also see directly that there are not unique solutions to the lifting problem. Let Cup be the morphism which goes from the empty 0-manifold to a pair of points, let Cap by the reversed morphism which goes from a pair of points to an empty 0-manifold. Let $\ell$ be the disjoint union of two Cups and $r$ be the disjoint union of two caps. $\ell \in L$ and $r \in R$. The composite $r \circ \ell$ is a disjoint union of two circles.

The target of $\ell$ is a 0-manifold which consists of two positive points and two negative points. Let $x$ be the invertible bordism which exchanges both the positive points and also exchanges the negative points. We have the identities:

$$ x \circ \ell = \ell $$ $$ r \circ x = r $$

So now we consider the lifting problem for the square where both sides are: do $\ell$ first, then do $r$. The identity bordism clearly gives a trivial solution to the lifting problem. However the bordism $x$ also gives a solution, so your classes do not form an orthogonal factorization system.

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  • $\begingroup$ Sorry @Chris, the $\mathcal{L}$ and $\mathcal{R}$ were mixed up above. I've fixed it, but unfortunately your answer no longer applies. Sorry my mistake wasted your nice explanation. $\endgroup$ – David Spivak Aug 3 '14 at 13:55
  • $\begingroup$ "the existence of a factorization follows by choosing a self-indexing Morse function (and dealing with the circle components separately)" I'm thinking about this since a long time: a certain OFS on Cob should describe (at least a piece of) Morse theory. Are you aware of somebody who tried to do this? $\endgroup$ – Fosco Loregian Aug 3 '14 at 17:48
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This question was answered in the affirmative by Joseph Abadi, here.

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