Let $C$ be a coalgebra and $\Delta: C \to C\otimes C$ a co-multiplication map. Then, due the co-associative property we can consider $\Delta^m$. But how is defined $\Delta^{m}: C \to C^{\otimes m}$?


Given $f,g \in C$ and $1\leq k \leq m$ can we have


$$\begin{align\*}\Delta^{m-1}(fg)&=\Delta^{k-1}(fg) \otimes \operatorname{id}^{m-k} + \Delta^{k-1}(f)\otimes \Delta^{m-k-1}(g) \\
&+\Delta^{k-1}(g)\otimes \Delta^{m-k-1}(f)+\operatorname{id}^{\otimes k} \otimes \Delta^{m-k-1}(fg)?\end{align\*}$$


Thanks.



How to work with co-multiplication?
To add to the answer above, I'd like to advertise the so-called *sumless Sweedler notation* here. 


This notation works as follows: let $C$ be a coalgebra; then if $c \in C$ then we write $\Delta(c) = c\_1 \otimes c\_2$ as an abbreviation of the more precise $\Delta(c) = \sum\_{i=1}^k c\_{i1} \otimes c\_{i2}$ for some $c\_{i1}, c\_{i2} \in C, i = 1,\ldots, k$. Then if we wish to consider $(1 \otimes \Delta) \circ \Delta : C \to C\otimes C \otimes C$ we simply write 


$$[(1 \otimes \Delta) \circ \Delta] (c) = c\_1 \otimes c\_2 \otimes c\_3.$$


In this notation, the coassociativity axiom $(1 \otimes \Delta) \circ \Delta = (\Delta \otimes 1) \circ \Delta$ becomes the inevitable 


$$ c\_1 \otimes (c\_2 \otimes c\_3) = (c\_1 \otimes c\_2) \otimes c\_3.$$


Sweedler's book "Hopf algebras", Susan Montgomery's book "Hopf algebras and their actions on rings" and the Wikipedia article <http://en.wikipedia.org/wiki/Coalgebra> are good references.


