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Corrected a typo in the formula
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Denis Nardin
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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)\supseteq \Sigma^n_T SH^{eff}(S)$$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

$f_{n+1}f_n = i_{n+1}r_{n+1}i_nr_n = i_n j_{n+1} l_{n+1} r_n i_n r_n = i_n j_{n+1}l_{n+1}r_n = f_{n+1}\,.$

The key here is that $SH^{eff}(S)$ is closed under suspensions, so there's an inclusion $j_{n+1}:\Sigma^{n+1}_T SH^{eff}(S)\supseteq \Sigma^n_T SH^{eff}(S)$. 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 adjoint of $j_{n+1}$. So

$f_{n+1}f_n = i_{n+1}r_{n+1}i_nr_n = i_n j_{n+1} l_{n+1} r_n i_n r_n = i_n j_{n+1}l_{n+1}r_n = f_{n+1}\,.$

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

$f_{n+1}f_n = i_{n+1}r_{n+1}i_nr_n = i_n j_{n+1} l_{n+1} r_n i_n r_n = i_n j_{n+1}l_{n+1}r_n = f_{n+1}\,.$

Source Link
Denis Nardin
  • 16.5k
  • 2
  • 69
  • 103

The key here is that $SH^{eff}(S)$ is closed under suspensions, so there's an inclusion $j_{n+1}:\Sigma^{n+1}_T SH^{eff}(S)\supseteq \Sigma^n_T SH^{eff}(S)$. 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 adjoint of $j_{n+1}$. So

$f_{n+1}f_n = i_{n+1}r_{n+1}i_nr_n = i_n j_{n+1} l_{n+1} r_n i_n r_n = i_n j_{n+1}l_{n+1}r_n = f_{n+1}\,.$