Isomorphism between sum of tensor products

51 Views Asked by At

In Sweedler book "Hopf algebras" (p.229) there is theorem that, if $(C_n)_{n\ge 0}$ is a filtration of coalgebra $C$ ( i.e. $C_0\subset C_1\subset \cdots\subset C, \ C=\bigcup_{n}C_n$ and $\Delta(C)\subset \sum C(i)\otimes C(n-i)$) then we can define associated graded coalgebra by $\operatorname{gr}C(n):=C_n/C_{n-1}, \ C_{-1}:=0$ with induced morphisms. In this theorem is used fact that

$ \left(\sum\limits_{i=0}^n C_i\otimes C_{n-i} \right) / \left(\sum\limits_{j=0}^{n-1}C_{j}\otimes C_{n-1-j}\right) \cong \sum\limits_{i=0}^{n}\left(C_i/C_{i-1}\right)\otimes\left(C_{n-i}/C_{n-i-1}\right)$

I tried to prove this statement by constructing a natural homomorphism

$\sum\limits_{i=0}^n C_i\otimes C_{n-i}\stackrel{\varphi}{\rightarrow}\sum\limits_{i=0}^{n}\left(C_i/C_{i-1}\right)\otimes\left(C_{n-i}/C_{n-i-1}\right)$

given by $\varphi \left(\sum\limits_{i=0}^n a_i\otimes b_{n-i}\right):=\sum\limits_{i=0}^n\overline{a_i}\otimes \overline{b_{n-i}}$

where $\overline{a_i}$ is image of $a_i$ given by natural epimorphism $C_i\rightarrow C_i/C_{i-1}$.

Maybe it is obvious, but I have problem with proving that $\ker \varphi $ is equal to $\sum\limits_{j=0}^{n-1}C_{j}\otimes C_{n-1-j} $. Any hints how to do it ? Is it a good way or should I prove it otherwise ?