# Zoran Skoda tensorF

$\array{ [ (\Delta \otimes_R id) F ] (F\otimes 1) \phi(1\otimes (\triangleright r) \otimes 1) & = & (F^1_{(1)}\otimes_R (F^1_{(2)}F'^1\triangleright r)\otimes F'2 \\ &=& \beta(F^1_{(2)}\triangleright(F'^2\triangleright r))F^1_{(1)} F'^2 \\ &=& \beta(F^1_{(2)}\triangleright (F'^2\triangleright r))F^1_{(1)}F'^1\otimes_R F^2 \\ &=& F^1\beta(F'^2\triangleright r)F'^1\otimes_R F^2 }$

where we have used $g\beta(b) = \beta(g_{(2)}\triangleright b) g_{(1)}$ for $g = F^1$. The identity

$[ (\Delta \otimes_R id) F ] (F\otimes 1) = [ (id \otimes_R \Delta) F ] (1\otimes F) \phi$

shows that the same can be calculated as

$\array{ [ (id \otimes_R \Delta) F ] (1\otimes F) \phi(1\otimes (\triangleright r) \otimes 1) & = & F^1 \otimes_R (F^2_{(1)} F'^{1}\otimes_R F^2_{(2)} F'^2) \phi (1 \otimes (\triangleright r)\otimes 1) \\ &=& F^1 \phi^1 \otimes_R \alpha (F^2_{(1)}\triangleright (F'^1 \phi^2 \triangleright a)) F^2_{(2)} F'^2 \\ &=& F^1\phi^1\otimes F^2\alpha(F'^1\phi^2\triangleright r)F'_2\phi^3 }$

where we have used $g\alpha(b) = \alpha(g_{(1)}\triangleright b)g_{(2)}$ for $g = F^2$.

We conclude that

$F^1\beta(F'^2\triangleright r) F'^1\otimes_R F^2 = F^1 \phi^1\otimes_R F^2\alpha(F'^1\phi^2\triangleright r)F'^2\phi^3$

for all $r$. In different terms,

$F I_F \subseteq I$

where $I = Ker(H\otimes H\to H\otimes_R H)$ and $I_F = Ker(H\otimes H\to H\otimes_F H)$ where $H\otimes_F H$ is the coequalizer of

$(H\otimes R)\otimes H \stackrel{\tilde\phi}\longrightarrow H\otimes(R\otimes H) \stackrel{H\otimes\alpha\otimes H}\longrightarrow H\otimes(H\otimes H)\stackrel{H\otimes \mu_H} \longrightarrow H\otimes H$

and

$(H\otimes R)\otimes H \stackrel{\tau_{R,H}\otimes H}\longrightarrow (R\otimes H)\otimes H \stackrel{(\beta\otimes H)\otimes H}\longrightarrow (H\otimes H)\otimes H\stackrel{\mu_H\otimes H} \longrightarrow H\otimes H$

Created on October 20, 2019 at 15:58:53. See the history of this page for a list of all contributions to it.