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I’m just here to check the code since there’s no way to check in the editor.
Proof: Take any isomorphism $f$, let $f = gh$ and $h f^{-1} = g' h'$ be the unique factorizations. Then $id = ghf^{−1} = (gg')h'$, so $h' = id$ and $gg' = id$, whence $g = id$ and $g' = id$ since $g, g' \in R_+$. Thus $f = h \in R_−$. The same argument applied to $f^{−1}$ shows that $f$ preserves the degree, hence $f = id$.
I’m just here to check the code since there’s no way to check in the editor.
Proof: Take any isomorphism $f$, let $f = gh$ and $h f^{-1} = g' h'$ be the unique factorizations. Then $id = gh f^{−1} = (gg')h'$, so $h' = id$ and $gg' = id$, whence $g = id$ and $g' = id$ since $g, g' \in R_+$. Thus $f = h \in R_−$. The same argument applied to $f^{−1}$ shows that $f$ preserves the degree, hence $f = id$.
For checking code for the $n$Lab best to use the Sandbox page.
The parser here on the $n$Forum does not behave identically to that for $n$Lab pages.
Added:
Monoidal Reedy model structures are discussed in
Added a reference to
I have re-organized and adjusted the section Definition – Plain version (here) to be more systematic.
(It used to start out with a rather pointless “Theorem” stating just that “there exists a model structure with objectwise weak equivalences”, then claimed to explain the “basic idea” but instead incrementally provided the actual definitions. I have turned all that around, added Definition/Proposition-environments etc. and pointers to the literature. )
In this comment, Charles Rezk once said that for $\mathcal{A}$ an additive model category, the Reedy cofibrations in $s \mathcal{A}$ correspond, under the Dold-Kan correspondence, exactly to the degreewise cofibrations in $Ch_{\geq 0}(\mathcal{A})$.
In trying to see this in detail, I am looking — given a morphism $f_\bullet \colon X_\bullet \to Y_\bullet$ in $s\mathcal{A}$ — at the diagrams
$\array{ L_r X &\longrightarrow& X_r \\ \big\downarrow && \big\downarrow \\ L_r Y & \longrightarrow & L_r Y \overset{L_r X}{\sqcup} X_r & \longrightarrow & Y_r \\ \big\downarrow && \big\downarrow && \big\downarrow \\ 0 &\longrightarrow& X_r / L_r X &\longrightarrow& Y_r/L_r Y }$Here the top square, the left rectangle and the bottom rectagle are pushouts by definition, whence the pasting law implies first that the left bottom square and then that the right bottom square is a pushout.
By pushout-stability of cofibrations, this yields that:
If $f_\bullet$ is Reedy cofibrant in that $L_r Y \overset{L_r X}{\sqcup} X_r \longrightarrow Y_r$ is a cofibration in $\mathcal{A}$ for each $r$
$\Rightarrow$ the induced morphism $X_r / L_r X \longrightarrow Y_r/L_r Y$ is a cofibration in $\mathcal{A}$, which is the $r$th component of the corresponding map of normalized chain complexes.
This gives one direction of the statement. But how about the backwards implication “$\Leftarrow$”?
Re #10: The other implication makes use of the concrete structure of the Dold–Kan functor Γ.
In detail, Γ will free degenerate elements in degree n along all possible degenerate simplicial maps.
This means that the latching object of ΓC in degree n can be computed simply as the inclusion of (ΓC)_n without C_n into (ΓC)_n.
This is an inclusion of a direct summand, therefore the latching map is a cofibration.
This is the argument that $L_r X \to X$ is a monomorphism in abelian categories. But the claim referred to in #10 is stronger, or even different, isn’t it?
Re #12: It’s not just monomorphism, it’s a split monomorphism, as shown in the construction of the Dold–Kan functor.
More precisely, the latching map for ΓC→ΓD will have the form L_n Y ⊔ C_n → L_n Y ⊔ D_n, so it is a cofibration as long as C_n → D_n is one, which it is by assumption.
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