Voevodsky defined the slice filtration on the motivic stable homotopy category $SH(S)$ over a Noetherian scheme $S$. In the article [Open Problems in the Motivic Stable Homotopy Theory, I](http://www.math.illinois.edu/K-theory/0392/newopen.pdf), Section 2, he defined $SH^{eff}(S)$ to be the smallest triangulated subcategory in $SH(S)$ which is closed under direct sums and contains suspension spectra of spaces. It follows from a result of Neeman that the inclusion functor 
$i\_n:\Sigma^n\_TSH^{eff}(S)\to SH^{eff}(S)$ has a right adjoint $r\_n$. Since $i\_n$ is full, we have a natural isomorphism $\mathrm{id}\simeq r\_n\circ i\_n$. Denote $f\_n=i\_n\circ r\_n$. Applying the counit $f\_{n+1}\to\mathrm{id}$ to $f\_n$, we get a natural transformation $f\_{n+1}\circ f\_n\to f\_n$. Voevodsky claims that $f\_{n+1}=f\_{n+1}\circ f\_n$ so that we get a natural transformation $f\_{n+1}\to f\_n$. The slice functor $s\_n$ is defined to be the cofiber of this map. Denoting the inclusion functor $\Sigma^n\_TSH^{eff}(S)\to \Sigma^{n-1}\_TSH^{eff}(S)$ by $j\_n$, (so that $i\_{n+1}=i\_n\circ j\_{n+1}$), I calculated that 
$$f\_{n+1}=i\_{n+1}\circ r\_{n+1}=i\_n\circ j\_{n+1}\circ r\_{n+1}=i\_n\circ\mathrm{id}\circ j\_{n+1}\circ r\_{n+1}$$
$$\simeq i\_n\circ r\_n\circ i\_n\circ j\_{n+1}\circ r\_{n+1}=f\_n\circ f\_{n+1}$$
but don't see why $f\_{n+1}=f\_{n+1}\circ f\_n$. Note that $f\_{n+1}=f\_n\circ f\_{n+1}$ also induces a natural transformation $f\_{n+1}\to f\_n$.


Am I missing something here? Or is there a typo in that paper?



Doubt regarding the definition of slice filtration
The key here is that $SH^{eff}(S)$ is closed under suspensions, so there's an inclusion $j\_{n+1}:\Sigma^{n+1}\_T SH^{eff}(S)\subseteq \Sigma^n\_T SH^{eff}(S)$. Hence you can write $i\_{n+1}=i\_n \circ j\_{n+1}$ and $r\_{n+1} = l\_{n+1}\circ r\_n$ where $l\_{n+1}$ is the right adjoint of $j\_{n+1}$ (which exists because $j\_{n+1}$ commutes with colimits). So


$f\_{n+1}f\_n = i\_{n+1}r\_{n+1}i\_nr\_n = i\_n j\_{n+1} l\_{n+1} r\_n i\_n r\_n = i\_n j\_{n+1}l\_{n+1}r\_n = f\_{n+1}\,.$


