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@Thorgott The following is a result of Hambleton & Kreck. Let $M, N$ be $4$-manifolds with finite fundamental group $G$, and suppose $M \#^k (S^2 \times S^2) \cong_{\mathrm{diff}} N \#^k (S^2 \times S^2)$. If $M = M_0 \# (S^2 \times S^2)$, then $M \cong_{\mathrm{top}} N$.
They prove this by using some results of Bass on cancellation of quadratic forms over $\Bbb ZG$. I have been wondering for some time if there is an alternative, simpler proof. Here is an attempt.
Consider the cobordism $M \Longrightarrow M \#^k (S^2 \times S^2)$ by gluing trivial $2$-handles, and a cobordism $N \Longrightarrow N \#^k (S^2 \times S^2)$ by gluing trivial $2$-handles. Then reversing the latter and pasting by the diffeomorphism in the hypothesis, one gets a cobordism $W : M \Longrightarrow N$.
The inclusion $M \to W$ is an isomorphism in $\pi_1$. I want to surger $W$ to make it a surjection on $\pi_2$. Note that for any space $X$, $\pi_2(X) = H_2(X; \Bbb Z\pi_1(X))$.
Say the cokernel of $\pi_2(M) \to \pi_2(W)$ is generated by some $2$-spheres. These can be ensured to be embedded since $\dim W = 5 > 2 + 2$. Suppose these have trivial normal bundle. Then I can obviously surger on them. The problem is this need not decrease the cokernel: surgery takes out $S^2 \times D^3$ and glues back $D^3 \times S^2$. The belt sphere $0 \times S^2$ may now be a new element in the cokernel.
Claim: Suppose $S \subset W$ is an embedded $2$-sphere such that $[S] \in H_2(W, M \sqcup N; \Bbb ZG)$ is not torsion. Then I can in fact surger $S$ out so that the rank of $\mathrm{coker}(\pi_2(M) \to \pi_2(W))$ decreases.
Proof: Suppose $T \subset W$ is the meridian of $S$, linking $S$ once. If I can show $[T] = 0 \in H_2(W \setminus S; \Bbb ZG)$, I am done -- then the belt sphere in fact is not giving rise to a new element in the cokernel after I surger along $S$.
To do this, consider the Poincare dual $\mathrm{PD}[S] \in H^2(W; \Bbb ZG) \cong H^2_c(\widetilde{W}; \Bbb Z)$. Since $W$ is finite fundamental group, cohomology with compact supports is actually just cohomology. So, $\mathrm{PD}[S] \in H^2(\widetilde{W}; \Bbb Z)$. Since $[S]$ is non-torsion, so is $\mathrm{PD}[S]$. Consider its universal coefficients dual in $H_2(\widetilde{W}; \Bbb Z)$, which is thus non-zero.
I can represent the resulting non-zero element of $H_2(\widetilde{W}; \Bbb Z)$ by a map $f : Y^3 \to \widetilde{W}$ from a closed $3$-stratifold $Y$ (in fact, since Steenrod realization is solvable in this dimension, $Y$ can be chosen to be a $3$-manifold). Composing with the covering map, I get $f : Y^3 \to W$ which intersects $S$ transversely with intersection number $+1$, because $[Y]$ represents PD of $[S]$.
Suppose $Y \pitchfork S = \{x_1, \cdots, x_n\}$. Consider $Y \setminus \cup_i B_\varepsilon(x_i)$. Note that $\partial B_\varepsilon(x_i)$ is either $T$ or $-T$ ($T$ with orientation reversed). So, $Y \setminus \cup_i B_\varepsilon(x_i)$ is a $3$-chain in $W \setminus S$ bounding $\sum \pm [T] = [T]$ (since the signs add up to the intersection number $+1$).
So $[T] = 0 \in H_2(W \setminus S; \Bbb ZG)$, as desired.
Suppose $H_2(W, M \sqcup N; \Bbb ZG)$ does not have torsion. I think I have a way of getting rid of torsion by doing the same thing as above with $\Bbb F_p$ and $\Bbb Q$ coefficients, but that will take us far.
So, after finitely many procedures of the above sort, I can ensure $H_2(W, M \sqcup N; \Bbb ZG) = 0$. One has to get from here to $H_2(W, M; \Bbb ZG) = 0$. This seems to require additional hypothesis. Let's get around to this later.
Remember
this? The $\Bbb ZG$-valued cellular chain complex from $(W, M)$ is $0 \to C_3(W, M; \Bbb ZG) \stackrel{\partial}{\to} C_2(W, M; \Bbb ZG) \to 0$ and if $H_2(W, M; \Bbb ZG) =0$, then $\partial$ is surjective. The kernel of $\partial$ is a stably free $\Bbb ZG$-module.
If stably free $\Bbb ZG$-modules were free and $\mathrm{Wh}(\Bbb ZG) = 0$, then essentially by the same arguments using elementary collapse/expansion moves, I can find a middle level set $\Sigma \subset W$ such that the sub-cobordism $W' : M \Longrightarrow \Sigma$ satisfies (1) $M \to W'$ is an iso on $\pi_1$, (2) $H_*(W', M; \Bbb ZG) = 0$. If also (3) $\Sigma \to W'$ is an iso on $\pi_1$, then $W'$ would be an $s$-cobordism.
The rest of $W$ after $W'$ has only some $3$-cells attached to $\Sigma$. Seen from the other side, $\Sigma$ is obtained from $N$ by attaching $2$-cells. But $N \to W$ is also an isomorphism on $\pi_1$, so the $2$-cells must be attached trivially. So $\Sigma \cong N \#^p (S^2 \times S^2)$. But if $W'$ is an $s$-cobordism, Freedman's theorem tells me $M$ and $N$ are homeomorphic. So $M \cong_{top} N \#^p (S^2 \times S^2)$. A Euler characteristic argument tells me $p = 0$, I think.
Anyway, "(3) $\Sigma \to W'$ is an iso on $\pi_1$" is not difficult to argue. $\pi_1 \Sigma = \pi_1 N$ by my description above. And $N \to W$ is an iso on $\pi_1$, so $\Sigma \to W'$ better be.
So it seems to me that in certain cases this is stronger than Hambleton-Kreck if I can ensure (0) Spheres in $W$ have trivial normal bundle (eg, $W$ is spin -- maybe) (1) $H_2(W; M \sqcup N; \Bbb ZG)$ does not have $\Bbb Z$-torsion (2) $H_2(W, M \sqcup N; \Bbb ZG) = 0$ implies $H_2(W, M; \Bbb ZG) = 0$, (3) Stably free $\Bbb ZG$-modules are free, (4) $Wh(\Bbb ZG) = 0$. Then $M \cong_{top} N$, without further reducibility assumption on $M$.
