Let $M\_1$ and $M\_2$ be two oriented, *connected*, closed $n$-manifolds. [It is known](http://www.map.mpim-bonn.mpg.de/Bordism#Connected_sum_and_bordism) that the disjoint union $M\_1 \sqcup M\_2$ and the connected sum $M\_1 \# M\_2$ are cobordant, via a bordism $\Sigma\_{M\_1, M\_2}$.


I'm wondering whether this bordism universal, in the following sense:



> 
> **Conjecture.** Let $N$ be any oriented, *connected*, closed $n$-manifold, and $\Sigma$ a bordism from $M\_1 \sqcup M\_2$ to $N$. Then there exists an essentially unique bordism $\Sigma'$ from $M\_1 \# M\_2$ to $N$ such that $\Sigma$ is the gluing of $\Sigma\_{M\_1, M\_2}$ an $\Sigma'$.
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**Is this true?**


I have a vague feeling that there must be an argument along the lines of shifting around certain critical points of a Morse function, and $\Sigma\_{M\_1, M\_2}$ corresponding to an extremal critical point, but I couldn't find a precise solution.


Edit: $\Sigma$ is required to be connected as well.



Is the bordism from disjoint union to connected sum universal for connected manifolds?
If we assume $\Sigma$ is connected, I think it's possible to prove existence, as [Mark Grant's counterexample](https://mathoverflow.net/a/261703) requires $\Sigma$ to be disconnected. The idea is that some disjoint-union-to-connect-sum bordism must appear, and then moving handles around turns it into $\Sigma\_{M\_1M\_2}$.


Decompose $\Sigma$ as a union of elementary cobordisms $\Sigma\_1,\dotsc,\Sigma\_k$. Since $\Sigma$ goes from a manifold with 2 connected components to a manifold with 1 connected component, one of these elementary cobordisms must go from a manifold with 2 connected components to a manifold with 1 connected component. Let $\Sigma\_i$ be the first one to do so, so that $\Sigma\_1,\dotsc,\Sigma\_{i-1}$ are disjoint unions of cobordisms applied to $M\_1$ and cobordisms applied to $M\_2$.


Let $P\_1\amalg P\_2$ denote the incoming boundary of $\Sigma\_i$. Since $\Sigma\_i$ is connected, it must be the trace of a surgery that removes an $S^0\times D^n$ and glues in a $D^1\times S^{n-1}$, which is the disjoint-union-to-connect-sum bordism $\Sigma\_{P\_1P\_2}$.


Now we need to get from $P\_1$ and $P\_2$ to $M\_1$ and $M\_2$. The part of $\Sigma$ before $\Sigma\_i$ is a disjoint union of two pieces, which are (without loss of generality) a cobordism $X\_1$ from $M\_1$ to $P\_1$ and a cobordism $X\_2$ from $M\_2$ to $P\_2$.


Now, $X\_1$ admits a handle presentation that begins with $M\_1\times[0,1]$ and attaches various handles. The diffeomorphism class of $\Sigma$ is unchanged if we instead attach these handles after $\Sigma\_i$ (if $\Sigma\_i$ is the final elementary bordism, append $N\times[0,1]$). The same argument applies to $X\_2$, so we've started with $\Sigma$, which is "($(M\_1\amalg M\_2)\times [0,1]$ plus some handles), then $\Sigma\_i$, then the rest" and obtained something diffeomorphic, which is "$(M\_1\amalg M\_2)\times[0,1]$, then $\Sigma\_{M\_1M\_2}$, then (the rest plus those handles)". Let $\Sigma'$ be "the rest plus those handles" and you have the desired decomposition.


