Graphical calculus for braided Hopf algebras I am trying to understand the graphical calculus presented in Ohtsuki's book Quantum Invariants. I think if I understand these first few examples it will help me greatly.
Let $(A,m,i,\Delta,\epsilon,S)$ be a Hopf algebra, where $m$ is multiplication, $i$ is the unit, $\Delta$ is comultiplication, $\epsilon$ is the counit, and $S$ is the antipode.
Q1: Two of the axioms that a Hopf algebra must satisfy are
$$
m \circ (S \otimes \text{id}) \circ \Delta = i \circ \epsilon,
$$
$$
m \circ (\text{id} \otimes S) \circ \Delta = i \circ \epsilon.
$$
In Ohtsuki's notation, these are written as

He states "each strand implies a copy of the Hopf algebra $A$. Further, along each strand we consider the product of the elements on the strand along the orientation of the strand." My question is: why are the strands coming from the bottom on the left and from the top on the right?
Q2: Consider a universal R-matrix $R \in A \otimes A$. There is one axiom that ensures that $m \otimes n \mapsto R(n \otimes m)$ is an $A$-module isomorphism $M \otimes N \to N \otimes M$: $(P \circ \Delta)(x) = R\Delta(x)R^{-1}$, where $P(x \otimes y) = y \otimes x$. The other two axioms turn $A$-mod into a braided tensor category:
$$
(\Delta \otimes \text{id})(R) = R_{13}R_{23},
$$
$$
(\text{id} \otimes \Delta)(R) = R_{13}R_{12}.
$$
Where $R_{12} = R \otimes 1$, $R_{23} = 1 \otimes R$, and $R_{13} = \sum \alpha_i \otimes 1 \otimes \beta_i$ putting $R = \sum \alpha_i \otimes \beta_i$. In Ohtsuki's notation, these are written as

However, this looks backwards to me. I read the diagram on the left as saying $(\Delta \otimes \text{id})(R) = R_{13}R_{12}$ and the diagram on the right as $(\text{id} \otimes \Delta)(R) = R_{13}R_{23}$. Or, possibly, $(\Delta \otimes \text{id})(R) = R_{12}R_{13}$ and $(\text{id} \otimes \Delta)(R) = R_{23}R_{13}$ if multiplication is performed in the other direction. I would think this is a typo but my intuition doesn't seem to work on any other diagrams later in this section (4.1). I also don't understand where all the permuting of the strands is coming from - every instance of $R$ is always preceded by a swap of the strands going into it, but that doesn't seem to be present in the definition of $R$ (except maybe the first axiom, but I don't see why that should mean we should precede every copy of $R$ by a swap of incoming strands). So my second question is to explain these diagrams.
I am also interested in any sources that explain this notation more thoroughly. There seems to be several variants of this kind of notation, but I'm not sure if this particular one has a name.
 A: Honestly, the presentation of the graphical calculus is rather mystifying to me, and I'm not sure I know another source of such diagrams, but I think I can offer some non-zero guidance by extrapolating from what is written.
For (Q1), I am not sure there is an intrinsic reason for having one be a cup and one a cap. I suspect that the reason has to do with the Hopf structure on tangles mentioned in the chapter preface. In particular, it seems to be the case that we want to change the direction of the strand before applying the antipode to an element (so up generally corresponds to $S(x)$, $S^3(x)$, etc. while down corresponds to $y$, $S^2(y)$, etc.). 
Since we put elements on the strands from left to right (i.e. a box containing $x\otimes y\otimes z$ means we place $x$ on the leftmost strand, $y$ on the middle, $z$ on the right, and multiply them in an order prescribed by the orientation), this more or less forces us to these diagrams. For example, in the first diagram we need summands of the form $S(x_1)x_2$, so necessarily we need one strand passing through the box twice oriented such that $S(x_1)$ appears first. Since we want to apply the antipode, we expect an upward strand, so we need a cap to come back to the box.
Is this satisfying? Probably not. I want to say that you can literally think of these diagrams as maps between tensor powers of $A$ with the boxes indicating left multiplication by the specified element, but that does not seem to be the case, which makes diagrammatic equalities such as those following (4.15) extremely puzzling. (What does a cap mean absent decoration? I don't know, but it makes sense when you put it back into the context of (4.15).)
(Q2) Okay, there are many questions we can ask about the meaning of diagrams here, but let's figure them out by writing things down carefully. Let's compute either side of (4.10), which is supposed to represent the fact that $R$ needs to change the coproduct to the transposed coproduct. First, note that explicitly, the diagram is supposed to reflect the identity
$$\sum_{i} x_2\alpha_i\otimes x_1\beta_i=\sum_{i} \alpha_ix_1\otimes \beta_i x_2.$$ 
Now, to decode (4.10), we follow strands along the orientation and multiply the corresponding elements. It is unclear from the text which tensor factor of $A\otimes A$ we land in, but we have our earlier equation as a guide.
On the left-hand side, following the strand at the top left and setting $\Delta(x)=\sum x_1\otimes x_2$, we get $x_1$ followed by $\beta_i$.
Comparing to our explicit identity, this seems to suggest that to compute the tensor factor corresponding to a downward-oriented strand at the bottom, we should read backwards along the strand adding new elements on the left to get the result. Note that this means a twist at the top doesn't effect the outcome
(since there is nothing left to multiply in that case). On the other hand, we can't do much interesting with only the $R$-matrix, so we pretty much exclusively use $R$ preceded by a twist.
So, again, on the left-hand side of (4.10), looking at the bottom left strand,
we encounter $\alpha_i$, then $x_2$. Since we read against the orientation, we multiply in the opposite order hence we get $x_2\alpha_i$ in the first slot.
Doing this with all the strands, we get 
$$LHS=\sum_i x_2\alpha_i\otimes x_1\beta_i,\quad RHS=\sum_{i} \alpha_ix_1\otimes \beta_i x_2.$$
Yay!
Now, do this with the identities you are worried about and things become quite clear. For instance, the left-hand side of the left-hand identity is clearly just $\Delta\otimes 1(R)$ since the twists at the top don't change anything, as argued earlier. On the right-hand side however, we can follow the strands to get
$$\sum \alpha_i\otimes \alpha_j\otimes \beta_i\beta_j=R_{13}E_{23}$$
(By the way, this does make sense as far as reading the diagram: it literally reads as $P_{12}R_{12}P_{23}R_{23}=P_{12}P_{23}R_{13}R_{23}$.)
Hopefully this can help you demystify things. (Though as a former French teacher of mine used to say, perhaps it only made things "clear as mud".)
