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In a strict monoidal category, where the associator, left and right unitor are identity morphisms we have the following relations between (string) diagrams:

(source)

where $i_x$ and $e_x$ are the unit and counit isomorphisms.

For the first relation for example, the LHS starts by $1\otimes x$ and goes to $x\otimes 1$, but the two are really the same object $x$ since we're in a strict monidal category, so we go from $x$ to $x$, which is exactly what happens in the RHS.

What I don't understand, is why such equalities are still valid in a weakened monoidal category, where now $x\otimes 1$ and $1\otimes x$ are merely isomorphic.

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The validity of string diagram equalities should be viewed as a form of coherence. What does an equality of two string diagrams tell us? Well, given such an equality, we can fix an arbitrary parenthesization and unitization of the input and the output. The associators and unitors are suppressed in a string diagram with the understanding that any two valid ways of adding them in give the same morphism. So in your example of the first zig-zag diagram, we can take both the input and output to be $x$, in which case the right-hand side is just the identity, and the left-hand side is given by the composite

$$x \stackrel{\lambda_x^{-1}}{\to} 1 \otimes x \stackrel{i_x \otimes x}{\to} (x \otimes x^{\vee}) \otimes x \stackrel{\alpha_{x, x^{\vee}, x}}{\to} x \otimes (x^{\vee} \otimes x) \stackrel{x \otimes e_x}{\to} x \otimes 1 \stackrel{\rho_x}{\to} x.$$

Here I've added associators and unitors where needed to get domains and codomains of morphisms to match up; I can do this with the confidence that had I chosen another way of adding associators and unitors, the composite morphism would be the same. This is what coherence tells us.

On the other hand, we could take the input to be $1 \otimes x$ and the output to be $x \otimes 1$. In this case, the right-hand side would represent the morphism $$1 \otimes x \stackrel{\lambda_x}{\to} x \stackrel{\rho_x^{-1}}{\to} x \otimes 1,$$ while the left-hand side would be given by $$1 \otimes x \stackrel{i_x \otimes x}{\to} (x \otimes x^{\vee}) \otimes x \stackrel{\alpha_{x, x^{\vee}, x}}{\to} x \otimes (x^{\vee} \otimes x) \stackrel{x \otimes e_x}{\to} x \otimes 1.$$

Of course, there are infinitely many possible choices of input and output, since I can just tensor 1 arbitrarily many times on the right and left and parenthesize this however I want. But once I fix a choice of input and output, each string diagram defines an unambiguous morphism.

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  • $\begingroup$ So if i understand correctly, associators and unitors are suppressed from the diagrams because, for example, to pass from $(x \otimes x) \otimes 1 \otimes (x \otimes 1 \otimes x)$ to $((x \otimes x) \otimes x) \otimes x$ there is only one way, one isomorphism, thanks to the coherence law. $\endgroup$
    – Pedro
    Commented Aug 8, 2012 at 18:58
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    $\begingroup$ @Pedro - this is correct, when the isomorphism is built from structure maps (associator, unitor etc). $\endgroup$
    – David Roberts ♦
    Commented Aug 13, 2012 at 4:22
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    $\begingroup$ Btw, string diagram are a way to describe compositions in 2-categories (see e.g. math.wisc.edu/~andreic/publications/Mukai1.pdf ). They are not a specificity of monoidal categories (which are nothing but 2-categories with a single object). $\endgroup$
    – DamienC
    Commented Aug 13, 2012 at 14:38

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