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$\DeclareMathOperator\Hom{Hom}\DeclareMathOperator\const{const}$Let $[+1]:\Delta\to\Delta$ be the décalage endomorphism sending $n\mapsto n+1$ adding a new minimal element, i.e. $f:n\to m$ is sent to $f':n+1\to m+1$, $f'(0)=0$, and for $0\leq i \leq n$ $f'(i+1)=f(i)+1$.

For $Y\in \operatorname{Ob} \mathit{sSets}$, let $Y^{\Delta[1]}$ denote the internal hom of maps of $\Delta[1]$ to $Y$.

Finally, let $\const(Y_0)$ denote the discrete object $n\mapsto Y_0$. Recall that taking the 0-face defines a map $Y\circ[+1]\to \const(Y_0)$ and it is a weak equivalence by (Lemma 1.5.1, K.Waldhausen, Waldhausen, Friedhelm Algebraic K-theory of spaces. Algebraic and geometric topology (New Brunswick, N.J., 1983), 318-419, Lecture Notes in Math., 1126, Springer, Berlin, 1985.) (thanks to @John Rognes's answer in Defining homotopy via endofunctors of a simplicial category ).

Is it true that

$Y\circ [+1]$ is isomorphic to $\const(Y_0)\times_Y Y^{\Delta[1]}$ ?

Here $\const(Y_0)\to Y$ is the obvious diagonal map, and $Y^{\Delta[1]\to Y$ is one of the two standard projections.

Sketch of possible proof: Let us construct a map $Y\circ [+1] \to Y^{\Delta[1]}$ which induced the isomorphism together with the map $Y\circ[+1]\to \const(Y_0)$.

Let $n$ denote the linear order $[0<1<\dotsb<n]$.

A map $(Y\circ [+1])_n=Y_{n+1}=\Hom(\Delta[n+1],Y) \to (Y^{\Delta[1]})_n=\Hom(\Delta[n]\times \Delta[1],Y)$ is given by a map $\Delta[n]\times \Delta[1]\to \Delta[n+1]$. To define this map, let us define for each $N$ a map $\Hom(N,n)\times \Hom(n,1) \to \Hom(N,n+1)$ by $(f,g)\mapsto (f+1)g$.

Note that when $g:N\to 1 $ sends everything to $0\in[0<1]$, the map $(f+1)g:N\to n+1$ sends everything to $0\in[0<1<\dotsb<n+1]$, and therefore $Y\circ[+1]\to Y^{\Delta[1]}\xrightarrow{0\in[0<1]} Y$ is the map $Y\circ[+1]\to \const(Y_0)\to Y$.

Note that when $g:N\to 1 $ sends everything to $1\in[0<1]$, the map $(f+1)g:N\to n+1$ sends each $i$ to $i+1$, and therefore $Y\circ[+1]\to Y^{\Delta[1]}\xrightarrow{1\in[0<1]} Y$ is the map $Y\circ[+1]\to Y$.

Does this argument really work ? Is there a reference for the fact it aims to prove ?

The motivation is that if the fact is true, then the property

a map $X\to Y$ factors though $Y\circ[+1]\to Y$

defines the class of maps homotopic to a map $X\to \const(Y_0)$, i.e. homotopically trivial on each connected component of the domain. And this property can be formulated in an arbitrary category of simplicial objects defining a class of almost homotopically trivial maps. I am also looking for references studying this property for an arbitrary category of simplicial objects.

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  • $\begingroup$ Writing $\mathrm{Dec}_+:= Y \circ [+1]$ as in Illusie's complexe cotangent (1972), the inclusion $i : Y_0 \to Y$ factorises as $Y_0 \to \mathrm{Dec}_+Y \to Y$. If $Y$ is a Kan complex, the first map is a weak equivalence and the second a Kan fibration. Thus a map $X \to Y$ factorises through $Y_0$ in the homotopy category iff it lifts to $ \mathrm{Dec}_+Y$. Although $Y_0\times_Y Y^{\Delta^1}$ is another space with this property, the final conclusion holds without needing any intermediate comparison; path objects aren't unique. $\endgroup$
    – Jon Pridham
    Commented Sep 1, 2022 at 20:27
  • $\begingroup$ @JonPridham: how do you conclude that "$X\to Y$ factorises though $Y_0$ in the homotopy category iff it lifts to $Dec_+ Y$" ? By a standard diagram chasing argument following directly from the axioms of a model category ? $\endgroup$
    – user420620
    Commented Sep 1, 2022 at 21:02
  • $\begingroup$ Essentially yes. For instance, since $Y_0 \to Y_0\times_YY^{\Delta^1}$ is a trivial cofibration, you get a weak equivalence $ Y_0\times_YY^{\Delta^1} \to \mathrm{Dec}_+Y$ over $Y$ by the lifting property. $\endgroup$
    – Jon Pridham
    Commented Sep 2, 2022 at 5:28
  • $\begingroup$ thanks, yes, this is clear. I was confused ... $\endgroup$
    – user420620
    Commented Sep 7, 2022 at 14:23

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