From the famous book: Monopole and three manifold, Kronheimer and Mrowka(<https://www.maths.ed.ac.uk/~v1ranick/papers/kronmrowka.pdf>). It is known that:
Let $Y$ be a closed oriented $3$ manifold, choosing a spinc structure $\mathfrak s$ and metric $g$ and a generic perturbation $p$, one can construct the monopole Floer homology groups:
$$\check{HM}\_\*(Y,\mathfrak s, g,p),~\hat{HM}\_\*(Y,\mathfrak s, g,p),~\overline{HM}\_\*(Y,\mathfrak s, g,p).$$
The groups are graded over a set $\mathbb J\_s$ admitting a $\mathbb Z$ action.( details are given in Section 20-22). We define the negative completions(Definition 23.1.3 of the book) by 
$$\check{HM}\_\bullet(Y,\mathfrak s, g,p),~\hat{HM}\_\bullet(Y,\mathfrak s, g,p),~\overline{HM}\_\bullet(Y,\mathfrak s, g,p).$$
If we want to consider all spinc
structures at the same time, we need to consider the completed monopole Floer homology
$$\check{HM}\_\bullet(M,F;\mathbb F)=\bigoplus\_\mathfrak s\check{HM}\_\bullet(M,F,\mathfrak;\mathbb F).$$ 


To show that these homology groups are independent of the metric and the perturbation, the authors gave a property: a cobordism between 3-manifolds gives rise to homomorphisms between
their monopole Floer homologies(see Section 23-26). They construct a homomorphism from $\check{HM}\_\bullet(Y,g\_1,p\_1)$ to $\check{HM}\_\bullet(Y',g\_2,p\_2)$, where there is a cobordism from $Y$ to $Y'$. 


**Q** I do not understand the two points below:


* Why the authors use the negative completion, where we need it?
* If we just want to show that the monopole Floer homology $\check{HM}\_\*(Y,\mathfrak s)$ is independent of the metric and perturbation, can we just using the trivial cobordism $[0,1]\times Y$ to show a homomorphism $\check{HM}\_\*(Y,\mathfrak s,g\_1,p\_1) \to \check{HM}\_\*(Y,\mathfrak s, g\_2,p\_2)$? The homomorphism is given by counting the number of solutions of the zero-dim moduli space $M([a\_1],W^\*,[b\_2])$, where $W^\*=(-\infty,0]\times Y\cup I\times Y\cup[1,\infty)\times Y$, and $[a\_1]$ and $[b\_2]$ are the critical points of $(Y,\mathfrak s,g\_1,p\_1)$ and $(Y,\mathfrak s,g\_2,p\_2)$ respectively. I think the arguments of Section23-25 also work before taking the negative completion .


*PS* Let $G\_\*$ be an abelian group graded by the set $\mathbb J$ equipped with a $\mathbb Z$-action. Let $O\_a(a\in A)$ be the set of free $\mathbb Z$-orbits in $\mathbb J$ and fix an element $j\_a\in O\_a$ for each $a$. Consider the subgroups 
$$G\_\*[n]=\bigoplus\_a\bigoplus\_{m\geq n} G\_{j\_a-m},$$
 which form a decreasing filtration of $G\_\*$. We define the negative completion of $G\_\*$ as the topological group $G\_\bullet\supset G\_\*$ obtained by completing with respect to this filtration.



Cobordism monopole Floer homology
The first bullet is definitely explained in the book! Surely around where it was introduced, it has to do with summing over all spin-c structures. We need to pass to the completion because the 4-manifold can have infinitely many spin-c structures that would need to be used. 


The second bullet, yes. In general we should not expect results concerning completions of a graded group to also hold for the uncompleted group. But here we consider the trivial cobordism, and spin-c structures on $[0,1]\times Y$ are the same as spin-c structure on $Y$, so no completion is needed in this situation. 


