How to work with co-multiplication? 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.
 A: We usually define $$\Delta^3=(\mathrm{id}_C\otimes\Delta)\circ\Delta,$$ and, more generally, $$\Delta^{k+1}=(\mathrm{id}_C^{k-1}\otimes\Delta)\circ\Delta^k.$$
(Your last formula does not make sense in a coalgebra: you seem to be multiplying elements of $C$; even in bialgebra, though, things like "$\Delta(fg)\otimes\mathrm{id}$" are pretty strange...)
A: When working with coalgebras I often find it helpful to dualize and to consider the corresponding situation for multiplication $\mu: A \otimes A \to A$ with an algebra $A$. 
In this point of view $\Delta^m$ corresponds to $\mu^m: A^{m+1} \to A$, given by 
$$\mu^m(a_0 \otimes ... \otimes a_m) = (a_0 ... a_{m-1}) a_m.$$
It's easy to see that this formula is equivalent to 
$$\mu^m = \mu \circ (\mu^{m-1} \otimes id_A).$$
Dualizing again yields 
$$\Delta^m = (\Delta^{m-1} \otimes id_C) \circ \Delta.$$
BTW: If $\Delta$ is coassociative this equals Mariano's formula 
$$\Delta^m = (id_C^{\otimes m-1} \otimes \Delta) \circ \Delta^{m-1}.$$
A: 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.
