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I'm not sure about the following result:

Theorem (?): Let $f_{\bullet,\bullet}: X_{\bullet,\bullet}\rightarrow Y_{\bullet,\bullet}$ a map of bisimplicial sets such that for any natural number $n$, $H_{\ast}(X_{\bullet,n},\mathbb{Z})\rightarrow H_{\ast}(Y_{\bullet,n},\mathbb{Z})$is an isomorphism, then
$H_{\ast}(DX_{\bullet,\bullet},\mathbb{Z})\rightarrow H_{\ast}(DY_{\bullet,\bullet},\mathbb{Z})$ is an isomorphism. Where $DX_{\bullet,\bullet}$ is the diagonal simplicial set.

I was wondering about the following proposition:

Let $f_{\bullet,\bullet}: X_{\bullet,\bullet}\rightarrow Y_{\bullet,\bullet}$ a map of bisimplicial sets such that for any natural number n, $H_{\ast}(X_{\bullet,n},\mathbb{Z})\rightarrow H_{\ast}(Y_{\bullet,n},\mathbb{Z})$ is surjective (resp. injective), then
$H_{\ast}(DX_{\bullet,\bullet},\mathbb{Z})\rightarrow H_{\ast}(DY_{\bullet,\bullet},\mathbb{Z})$ is surjective (resp. injective).

Since I don't know how to prove the first theorem (?), I do not know if the second proposition has any chance to be correct? Are there any similar results in the literature?

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    $\begingroup$ For Thm. 1: the homology of the geom. realization of a simplicial set agrees with the homology of the alternating sign chain complex of the levelwise free abelian group on that simp. set. So you can use the spectral sequence of a double complex for your first theorem: the homology groups of of X_{\bullet, n}, across all n, form the input, while the homology of DX_{\bullet,\bullet} forms the output. This is old Cartan-Eilenberg stuff. Finally, you can use Boardman's observation, that if a map of spect. seq's induces an iso at some finite page, it induces an iso at all later pages. That does it! $\endgroup$
    – user124192
    Commented Jun 3, 2018 at 18:50
  • $\begingroup$ Thinking about the spectral sequence, and whether having a surjective map of spectral sequence E_2-terms implies that the map on E_3-terms is surjective, also gets you to an answer to your question about surjectivity alone (and injectivity alone, too). $\endgroup$
    – user124192
    Commented Jun 3, 2018 at 18:54

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Yes, this is true. I will suppress $\mathbb{Z}$ from the notation, this works with any coefficients or indeed with any homology theory.

We have isomorphisms $H_*(B, A) \cong \tilde H_*(B/A) \cong \tilde H_{*+k}(\Sigma^k B/A) \cong H_{*+k}(B \times \Delta[k], A \times \Delta[k] \cup B \times \partial\Delta[k])$. Thus if $A \hookrightarrow B$ is a homology equivalence, then so is $A \times \Delta[k] \cup B \times \partial\Delta[k]\hookrightarrow B \times \Delta[k]$.

From here you can use any of the two standard arguments for weak homotopy equivalences.

Goerss, Jardine (Proposition IV.1.9) provide an explicit filtration of the diagonal with each step obtained from the previous one by pushing out a pushout product map as above.

Alternatively, Hovey's proof (Lemma 5.3.1) also applies verbatim. It shows that the diagonal functor is left Quillen with respect to the Reedy model structure (starting with the homological Bousfield localization of course). Indeed, its right adjoint is $K \mapsto \mathsf{sSet}(\Delta[-],K)$ and it is right Quillen by the adjoint form of the pushout product property above (sometimes called the pullback hom property).

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  • $\begingroup$ This is for the first statement only, right? (I.e., not the one about surjective or injective maps only, which I wouldn't expect to be true) $\endgroup$
    – Achim Krause
    Commented Jun 4, 2018 at 17:26
  • $\begingroup$ Oh yes, I don't have a counterexample in mind, but I would be very surprised if this worked for surjections or injections. $\endgroup$
    – Karol Szumiło
    Commented Jun 4, 2018 at 18:52
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    $\begingroup$ It is easy to find maps between simplicial sets that are level-wise injective (resp. surjective), but such that the induced map on homology is not injective (resp. surjective). For example, think of inclusion of the boundary of an interval (resp. the quotient map from an interval to the circle). Thinking of these simplicial sets as bisimplicial sets that are constant in one direction gives the desired counterexamples. $\endgroup$
    – Gregory Arone
    Commented Jun 5, 2018 at 5:11

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