The tensor product of two topological complexes with closed range A Künneth formula by Grothendieck/Schwartz states the following:

Let $A, B$ be chain complexes of nuclear Fréchet spaces. If the differentials $d_A, d_B$ are topological homomorphisms (meaning in this setting: if they have closed ranges), then we have the Künneth formula $$H(A  \, \hat \otimes  \, B ) \cong H(A) \, \hat \otimes \, H(B). \quad \quad\quad (*)$$

I would like to apply this statement to more than two product factors, and to do so, it would suffice that the range of the differential on the product space be closed again. Is there an easy way to see why/if this is true? I've tried to deduce it from the proof given in their paper, but somehow I don't see it, because the range of the product differential does not seem to come up explicitly.
On the one hand, the closed range property seems quite fickle and is in general not closed under linear combinations, and I fear that an expression like $d_A \otimes \text{id} + \text{id} \otimes d_B$ might be too general to hope for the closed range property again.
On the other hand, the RHS of $(*)$ is canonically a Fréchet space again, the LHS only if the range of the product differential is closed. If this isomorphism held, but the product differential range was not closed, I feel that would be quite strange...
 A: The discussion in the comments catapulted me onto the right track! It seems the solution is exactly to note that the isomorphism $H(A  \, \hat \otimes  \, B ) \cong H(A) \, \hat \otimes \, H(B)$ is not only an isomorphism of abstract vector spaces, but indeed an isomorphism of topological vector spaces, where all homology spaces $\ker d/\text{im }d$ are equipped with the respective quotient topologies. If this is done, then the right-hand-side is a Fréchet space, so the left-hand-side is, too, and the quotient topology of a Fréchet space by a subspace if Fréchet if and only if the subspace was closed.
Proving this takes a bit more effort than what is done in the Schwartz/Grothendieck paper, but the paper "A Künneth formula in topological homology and its applications to the simplicial cohomology of $ \ell^1 (\mathbb{Z}_+^{k}) $." by Gourdeau, Lykova and White, mentioned by Yemon Choi in the comments, deals with this. They shows the precise statement in Corollary 5.3, under the assumption that $A$ and $B$ are bounded from below as complexes.
