Is the bordism from disjoint union to connected sum universal for connected manifolds? Let $M_1$ and $M_2$ be two oriented, connected, closed $n$-manifolds. It is known 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 $\Sigma'$.

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.
 A: I think the idea in Thomas Rot's answer (now deleted) can be modified to give a counterexample when $M_1$, $M_2$ and $N$ are connected.
Namely, assume $M_1$, $M_2$ and $N$ are all null-cobordant, and let $\Sigma$ be the disjoint union of three such null-cobordisms. Then there is no path from $M_1$ to $M_2$ in $\Sigma$, but there would be in a composition of $\Sigma_{M_1,M_2}$ with $\Sigma'$.
A: TLDR: existence yes, uniqueness no.
Arun has already discussed existence, so let me discuss the uniqueness part of the question.  It is a special case of whether cancellation holds in bordism categories, which it usually doesn't except under rather strong assumptions.
The uniqueness part of the question is equivalent to whether all embeddings $\Sigma_{M_1,M_2} \hookrightarrow \Sigma$ relative to $M_1 \amalg M_2$ are in the same orbit under the action of the group of diffeomorphisms of $\Sigma$ restricting to the identity on $\partial \Sigma$.  (One direction: given two such embeddings, their complements would be two candidates for $\Sigma'$ and any diffeomorphism rel boundary between these two complements may be glued to the identity map of $\Sigma_{M_1,M_2}$ to get a diffeomorphism of $\Sigma$ which acts one embedding into the other.  The converse is similar.)  In fact we may construct an example of two embeddings $e, e': \Sigma_{M_1,M_2} \to \Sigma$ where it is not even possible to find a continuous map $f: \Sigma \to \Sigma$ which is the identity on $\partial \Sigma$ and such that $f \circ e \simeq e'$ relative to $\partial \Sigma$.
To build a counterexample we use that for any group $G$, the set of $G \times G^\mathrm{op}$-equivariant maps $G \to G$ (action by multiplying on both sides) is precisely the maps given by multiplication by elements of the center of $G$.  Hence if $Z(G) = \{1\}$ it consists of only the identity.  This implies the following more complicated looking (but "isomorphic") statement: if $C$ is a groupoid and $m_1$ and $m_2$ are two isomorphic objects in $C$ such that the group $\mathrm{End}_C(m_1)$ has trivial center, then any functor $F: C \to C$ with $F(m_1) = m_1$ and $f(m_2) = m_2$, and $F(f) = f$ for any $f \in \mathrm{End}_C(m_1)$ and any $f \in \mathrm{End}_C(m_2)$, also has $F(f) = f$ for any $f \in \mathrm{Hom}_C(m_1,m_2)$.
Now pick a finitely presented non-trivial group $G$ with trivial center and build a connected oriented cobordism $W$ from $M_1 \amalg M_2$ to any connected $N$, such that there exists an isomorphism $\pi_1(W,w) \cong G$ for some, hence any, $w \in W$, and such that both inclusions $M_1 \to W$ and $M_2 \to W$ induce isomorphisms in fundamental groups.  For example, we could first pick a closed connected oriented manifold $M$ with the right fundamental group (this is possible in any dimension $d \geq 4$), then set $M_2 = M_1 = M$ (with opposite orientations), and let $\Sigma$ be the complement of a ball in the interior of $[0,1] \times M$, regarded as a bordism from $M_1 \amalg M_2$ to $N = S^{d-1}$.
Then pick points $m_1 \in M_1$ and $m_2 \in M_2$, which we can think of as objects in the fundamental groupoid $\pi_1(\Sigma)$.  Connectedness implies that these objects are isomorphic, and we have arranged that the group of endomorphisms of $m_1$ has trivial center.  By what we saw above, any functor $F: \pi_1(\Sigma) \to \pi_1(\Sigma)$ which sends both objects $m_1$ and $m_2$ to themselves and act as the identity on their endomorphism groups, must also act as the identity on the set of isomorphisms from $m_1$ to $m_2$.
The bordism $\Sigma_{M_1,M_2}$ may be constructed by attaching a 1-handle to $M_1 \amalg M_2$ at the points $m_1$ and $m_2$.  Then any embedding $\Sigma_{M_1,M_2} \to \Sigma$ gives rise to an isomorphism (namely the path through the attached 1-handle) from $m_1$ to $m_2$ in the fundamental groupoid of $\Sigma$, and any element of this Hom-set arises in this way (since any path may be perturbed to a smooth embedding, which by the orientability assumptions may be thickened up to an embedding of the handle).  Then pick two embeddings $e, e': \Sigma_{M_1,M_2} \to \Sigma$ representing distinct elements in the Hom-set.  Any diffeomorphism of $\Sigma$ relative to $M_1 \amalg M_2$ which acts $e$ into $e'$ would induce a functor $\pi_1(\Sigma) \to \pi_1(\Sigma)$ acting as the identity on the endomorphisms of $m_1$ and $m_2$ (since these endomorphism groups may be calculated in $M_1$ and $M_2$) but not on the set of morphisms between them.  But that's precisely what we ruled out!  Hence such a diffeomorphism cannot exist.
A: If we assume $\Sigma$ is connected, I think it's possible to prove existence, as Mark Grant's counterexample 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.
