Gluing a manifold along its boundary, via chain complexes Given closed oriented $n$-manifolds $M, M', M''$ and bordisms $W, W'$ with $\partial W = M \sqcup - M'$ and $\partial W' = M' \sqcup - M''$, we can collar-glue them to obtain a bordism from $M$ to $M''$. There is a corresponding statement on the level of the respective (singular) chain complexes:
Because of Poincaré Duality, $C^*(M)$ (we always choose coefficients in $\mathbb{Z}$ but other rings would also work) is equipped with a natural quasi-isomorphism $\omega$ to the shift of its dual chain complex
$$\omega: C^*(M) \simeq C^*(M)^\vee [-n]$$
induced by capping with the fundamental class, and similarly for $C^*(M')$ and $C^*(M'')$, let us call chain complexes equipped with such a quasi-isomorphism $n$-dimensional Poincaré complexes.
The bordism $W$ induces a so-called Lagrangian correspondence $C^*(M) \leftarrow C^*(W) \to C^*(M')$ where the morphisms are given by restrictions; this means that the relative fundamental class of $W$ induces a quasi-isomorphism
$$ \operatorname{fib}(C^*(W) \to C^*(M) ) \simeq \operatorname{fib} (C^*(W) \to C^*(M'))^\vee [-n-1]$$
where $\operatorname{fib}$ should denote the mapping fiber of a map of chain complexes (up to a shift by $1$, the mapping cone). Equivalently, one could also write
$$ C^*(W) \simeq \left( C^*(M) \amalg^h_{C^*(W)} C^*(M') \right)^\vee [-n] $$
where $\amalg^h$ denotes the homotopy pushout in chain complexes (the pushout in the corresponding $\infty$-category. See e.g. the beginning of Chapter 2 in this paper for a precise reference; on the level of cohomology groups, this is just Poincaré-Lefschetz Duality.
Now, one can show (using a pasting argument) that the Lagrangian correspondences $C^*(M) \leftarrow C^*(W) \to C^*(M')$ and $C^*(M') \leftarrow C^*(W') \to C^*(M'')$ can be composed to a Lagrangian correspondence $C^*(M) \leftarrow C^*(W) \times_{C^*(M')} C^*(W') \to C^*(M'')$ where the middle complex should be quasi-isomorphic to $C^*(W \cup_{M'} W')$ by excision. In other words, collar-gluing of bordisms has an algebraic analogon for Lagrangian correspondences.
My question: Given a compact oriented $(n+1)$-manifold $W$ with $\partial W = M \sqcup - M$, for $M$ a closed oriented $n$-manifold. Then, we can collar-glue $W$ to itself along $M$. For example if $W = S^1 \times [0,1]$ and $M = S^1$, we otain the torus. Is there also an algebraic version of this result, of the following form?
Let $(C, \omega)$ be an $n$-dimensional Poincaré complex and $C \overset{f}{\leftarrow} D \overset{f'}{\to} C$ be a Lagrangian correspondence with corresponding quasi-isomorphism
$$ \eta: \operatorname{fib} (f) \simeq \operatorname{fib} (f')^\vee [-n-1] \; .$$
Is the homotopy-equalizer $\operatorname{eq}(f,f') = \operatorname{fib}(f-f')$ an $(n+1)$-Poincaré-complex in a natural way? Does this result generalize to Poincaré objects in an arbitrary Poincaré $\infty$-category? I do not see how to prove this, and the case $f = f'$ seems dubious, so maybe one needs some extra conditions.
 A: I might have a idea how to prove my claim, that also generalizes to any stable $\infty$-category with duality functor:
Let $C$ be a chain complex, remember that there are natural diagonal and codiagonal chain maps $\Delta: C \to C \oplus C$ and $\nabla: C \oplus C \to C$, and for chain maps $f,g : D \to C$ the composition
$$ D \overset{\Delta}{\to} D \oplus D \overset{f \oplus g}{\to} C \oplus C \overset{\nabla}{\to} C $$
agrees with the sum $f + g$.
Since we assume $C \overset{f}{\leftarrow} D \overset{f'}{\to} C$ is a Lagrangian correspondence, we have
$$ \operatorname{fib}(f) \simeq \operatorname{fib}(f')^\vee [-n-1] \\
\operatorname{fib}(f') \simeq \operatorname{fib}(f)^\vee [-n-1] $$
where the second equation follows from the first by applying $(-)^\vee [-n-1]$ to both sides. Now, we take the direct sum of these equations and take homotopy invariants with respect to the $S_2$-action that exchanges direct summands:
$$(\operatorname{fib}(f) \oplus \operatorname{fib}(f'))^{hS_2} \simeq (\operatorname{fib}(f') \oplus \operatorname{fib}(f))^{\vee hS_2} [-n-1] $$
We can write the fiber on the right side as a shift of a cofiber, and pull the homotopy (co-)invariants inside:
$$\operatorname{fib} \left( (D \oplus D)^{hS_2} \overset{(f \oplus f')^{hS_2}}{\longrightarrow} (C \oplus C)^{hS_2} \right)  \simeq \operatorname{cofib} \left( (D \oplus D)_{hS_2} \overset{(f' \oplus f)_{hS_2}}{\longrightarrow} (C \oplus C)_{hS_2} \right) ^{\vee} [-n] $$
Now, we use the statement from the beginning combined with the fact that the diagonal map induces an isomorphism $C \simeq (C \oplus C)^{hS_2}$, just as the codiagonal induces an isomorphism $(C \oplus C)_{hS_2} \simeq C$:
$$\operatorname{fib}(D \overset{f + f'}{\longrightarrow} C)  \simeq \operatorname{cofib}(D \overset{f + f'}{\longrightarrow} C)^\vee [-n] \simeq \operatorname{fib}(f+f')^\vee [-n-1] $$
Up to exchanging $f'$ and $-f'$ which gives isomorphic Poincaré complexes (and is probably due to the different orientations in the manifold case), we have succeed since we have a quasi-isomorphism that exhibits $\operatorname{fib}(f+f')$ as an $(n+1)$-dimensional Poincaré compex.
Open Questions (assuming by argument is correct):

*

*Does $\operatorname{fib}(f-f')$ agree with the singular chain complex of the glued manifold, in the manifold case? This should again use excision, but Lurie's generalized Seifert-van-Kampen theorem is not applicable since the equalizer diagram does not have a weakly contractible classifying space.

*Figuring out the case $f = f'$ as an example.

