This question only concerns the final part of the proof, so I assume that the symmetric monoidal category is a strict monoidal category $\mathsf{C}$ with the braiding $s$.

Let $X_1,...,X_n$ be elements of $\mathsf{C}$ and let $\sigma \in S_n$. Mac Lane proves that any two morphisms $f\colon X_1\otimes ... \otimes X_n \to X_{\sigma(1)}\otimes ... \otimes X_{\sigma(n)}$ which are compositions of braidings possibly tensored with identity morphism on the left or on the right $k$-times are equal. To this end, he realizes every such path a composition $s^{\pm}_{X_{i_1}, X_{i_1 + 1}} \circ ... \circ s^{\pm}_{X_{i_m}, X_{i_m + 1}}$ (I ignore $-\otimes 1_X$ and $1_X\otimes -$ here as far as the notation goes). The key part of the proof in an apparently intereseting connection between $s_{X_i,X_i + 1}$ and $(i,i+1)$.

Mac Lane states and any "closed path" (I suppose he's refering to said morphism with codomain $X_1\otimes ... \otimes X_n$) corresponds to a relations between generators $(i,i+1)$ of the symmetric group $S_n$. On the other hand, he mentions that a symmetric group has the presentation $$\langle \tau_i, i = 1,...n-1 \mid \tau^2_i = 1, (\tau_i\tau_{i+1})^3 = 1, \tau_i\tau_j = \tau_j\tau_i \text{ for }|i - j| > 1 \}$$

where $\tau_i = (i,i+1)$. He then claims that to prove the statement it suffices to show that these relations hold for $s$.

I've been thinking for a while about this and I can't understand what is the precise connection between braidings and permutations. I see that applying a braiding $s_{X_i,X_{i + 1}}$ to $X_1\otimes ... \otimes X_n$ given b$X_{\sigma(1)} \otimes ... \otimes X_{\sigma(n)}$ where $\sigma = (i,i+1)$, but no better than that yet.

Also, connecting relations in $S_n$ with those among $s_{X,Y}$ and deducing the statement of the theorem from that reminds me of the universal property of a presentation:

Let $\langle X \mid R \rangle$ be a presentation and $G$ a group. Let $f\colon X\to G$ be a map such that for every $(u,v) \in R$ we have $f(u) = f(v)$. Then there is a unique group homomorphism $\phi\colon\langle X \mid R \rangle \to G$ such that $\phi(x) = f(x)$ for all $x \in X$.

But I don't understand how this can be applied here as I don't see a resonable binary operation for paths consisting of braidings.

So what Mac Lane really means here, and how it can be made precise?


I think that this is a great question, since it is quite common to just say at this point "each relation between the transpositions follows from the relations in the group presentation of the symmetric group, so it is enough to prove the relations for the symmetry automorphisms", but as you correctly point out, this is not really precise! One can make it precise by saying more about what a deduction of $a=b$ between two words in a group presentation actually is, and how in our case every such deduction can be transported to a deduction of an equality between the associated isomorphisms between the tensor products. But I won't go into that here, since I have actually found a cute trick.

Consider the special case $X_1=\dotsc=X_n$ first, let's call this unique object $X$. Then the task is to define a (well-defined) map $$S_n \to \mathrm{Aut}(X^{\otimes n}),\, \sigma \mapsto \sigma_*$$ which will actually be constructed as a homomorphism. By the presentation $$S_n = \langle \tau_1,\dotsc,\tau_n : \tau_i^2=1,\, (\tau_i \tau_{i+1})^3=1, \,\tau_i \tau_j = \tau_j \tau_i \text{ if } |i-j|>1 \rangle$$ it is therefore enough to define elements $s_i = (\tau_i)_* \in \mathrm{Aut}(X^{\otimes n})$ satisfying the mentioned relations. This is what Mac Lane actually proves.

The general case can be deduced from the special case:

Consider $\mathcal{C}_{\sqcup}$, the cocompletion of $\mathcal{C}$ under finite coproducts. Objects are finite (formal) coproducts $\coprod_{i \in T} X_i$ with $X_i \in \mathcal{C}$ and finite sets $T$, the morphisms are $$\hom\bigl(\coprod_{i \in T} X_i,\coprod_{j \in S} Y_j\bigr) = \prod_{i \in T} \coprod_{j \in S} \hom(X_i,Y_j).$$ There is an evident symmetric monoidal structrue on $\mathcal{C}_{\sqcup}$ extending the one on $\mathcal{C}$, which commutes with finite coproducts in each variable, thus given by $$\coprod_{i \in T} X_i \otimes \coprod_{j \in S} Y_j := \coprod_{(i,j) \times T \times S} X_i \otimes Y_j.$$ The symmetry $\coprod_{i \in T} X_i \otimes \coprod_{j \in S} Y_j \to \coprod_{j \in S} Y_j \otimes \coprod_{i \in T} X_i$ is induced by the symmetries $X_i \otimes Y_j \to Y_j \otimes X_i$. All coherence axioms immediately follow from the ones in $\mathcal{C}$.

Now consider for $X_1,\dotsc,X_n \in \mathcal{C}$ the object $X := X_1 \sqcup \dotsc \sqcup X_n$ in $\mathcal{C}_{\sqcup}$ with the coproduct inclusions $\iota_1,\dotsc,\iota_n$. We know that every $\sigma \in S_n$ induces a well-defined automorphism $X^{\otimes n} \to X^{\otimes n}$ in $\mathcal{C}_{\sqcup}$. Composing this with $\iota_1 \otimes \dotsc \otimes \iota_n$ yields a well-defined morphism $X_1 \otimes \dotsc \otimes X_n \to X^{\otimes n}$, still in $\mathcal{C}_{\sqcup}$. But if we write $\sigma$ as a product of neighbor transpositions and hence write $X^{\otimes n} \to X^{\otimes n}$ as a product of symmetry automorphisms, we see that the morphism actually factors over $X_{\sigma^{-1}(1)} \otimes \dotsc \otimes X_{\sigma{-1}(n)}$. So we get a morphism $X_1 \otimes \dotsc \otimes X_n \to X_{\sigma^{-1}(1)} \otimes \dotsc \otimes X_{\sigma^{-1}(n)}$ in $\mathcal{C}$. Once again, it is independent from any product decomposition of $\sigma$ since $X^{\otimes n} \to X^{\otimes n}$ is.

This reduction can also be informally phrased as follows: The objects are in their right place anyway, so we might as well just pretend that they are equal.

  • $\begingroup$ Thank you for your attention. If you don't mind, a I have a couple of questions. 1) Do "formal coproducts" mean families $(X_i)_{i \in I}$ for $I$ finite? I assume they will be real coproducts in $C_{\sqcup}$? 2) The composition of $(i_j, f\colon X_i\to Y_{j_i})_{i \in I} \in \prod_{i \in I} \coprod_{j \in J} \mathrm{Hom}(X_i,Y_j)$ and $(k_j, g\colon Y_j\to Z_{k_j})_{j \in J} \in \prod_{j \in J} \coprod_{k \in K} \mathrm{Hom}(Y_j,Z_k)$ is $(i, k_{j_i})$, right? The last request is the one I'm most uncomfortable writing, but could you, please, provide more details as to why the morphism $\endgroup$ – Jxt921 Jan 12 '20 at 20:56
  • $\begingroup$ $X_1\otimes ... \otimes X_n \to X^{\otimes n}$ factors over $X_{\sigma(1)}\otimes ... \otimes X_{\sigma(n)}$, and why it would imply the uniqueness? $\endgroup$ – Jxt921 Jan 12 '20 at 20:57
  • $\begingroup$ Sorry, in the question about the composition it should read as: $(j_i,f_i\colon X_i\to X_j)_{i \in I}$ and $(i,g_{j_i}\circ f_i)$ instead of $(i_j, f\colon X_i\to Y_{j_i})$ and $(i,k_{j_i})$. $\endgroup$ – Jxt921 Jan 12 '20 at 21:12
  • $\begingroup$ 1) Yes, 2) Yes, $\endgroup$ – Martin Brandenburg Jan 12 '20 at 21:24
  • $\begingroup$ 3) I just give you an example to get a better feeling. Let $\sigma = (1\,2\,3)$, thus $\sigma = (1\,2) \circ (2\,3)$ for example. Then $\sigma_* :X^{\otimes 3} \to X^{\otimes 3}$ is given by $s_1 \circ s_2$, where $s_1 = S_{X,X} \otimes X$ and $s_2 = X \otimes S_{X,X}$. If $X = X_1 \sqcup X_2$, then $s_2$ maps $X_1 \otimes X_2 \otimes X_3$ to $X_1 \otimes X_3 \otimes X_2$, and $s_1$ maps $X_1 \otimes X_3 \otimes X_2$ to $X_3 \otimes X_1 \otimes X_2$, which is actually $X_{\sigma^{-1}(1)} \otimes \dotsc \otimes X_{\sigma^{-1}(n)}$. Ok I will correct my answer. $\endgroup$ – Martin Brandenburg Jan 12 '20 at 21:25

I'm not sure, but it appears that it is not as hard as it seemed. Martin's answer is interesting, but I still wanted to try something elementary before fully committing to powerful machinery.

Let $F$ be the free group on $\{ \tau_i \mid i = 1,...,n - 1 \}$ where $\tau_i = (i,i+1)$ and let $\phi\colon F\to S_n$ be a canonical homomorphism where $S_n = F/N$ with $N$ is the normal closure of the subset of $F$ consisting of elements $\tau_i\tau_i, (\tau_i\tau_{i+1})^3$ and $\tau_i\tau_j\tau^{-1}_i\tau^{-1}_j$ (for $|i - j| > 1$).

Let $M$ be the set of all morphisms with domain $X_1\otimes ... \otimes X_n$. We define a map $f\colon F\to M$ given by $f(e) = 1_{X_1\otimes ... \otimes X_n}$ and $f(\tau^{\pm}_{i_n},...,\tau^{\pm}_{i_1}) = s_{X_{\tau_{i_1}\left(...\left(\tau_{i_{n-1}}(i_n)\right)\right)}, X_{\tau_{i_1}\left(...\left(\tau_{i_{n-1}}(i_n + 1)\right)\right)}}\circ...\circ s_{X_{\tau_{i_1}(i_2)}, X_{\tau_{i_1}(i_2 + 1)}}s_{X_{i_1},X_{i_1 + 1}}$ which is a morphism $X_1\otimes ... \otimes X_n \to X_{\sigma^{-1}(1)}\otimes ... \otimes X_{\sigma^{-1}(n)}$ with $\sigma = \tau_{i_n},...,\tau_{i_1}$. Since coherence states that only all "formal" diagrams commute, every path $X_1\otimes ... \otimes X_n \to X_{\sigma^{-1}(1)}\otimes ... \otimes X_{\sigma^{-1}(n)}$ consisting of symmetries is of this form.

If a path of this form from $X_1\otimes ... \otimes X_n \to X_1\otimes ... \otimes X_n$, then $\sigma = 1_{\{1,...,n\}}$, so $\tau^{\pm}_{i_n},...,\tau^{\pm}_{i_1} = g_1n_1g^{-1}_1...g_mn_mg^{-1}_m \in N$

Now the problem is that $g_1n_1g^{-1}_1...g_mn_mg^{-1}_m$ is not necessarily reduced, so it's possible that it doesn't translate to a product of conjugates of $n_1,...,n_m$ in $M$. We only need to note that, for example, $f(\tau_{i_1}\tau_{i_1}^{-1}) = s_{X_{\tau_{i_1}(i_1)}, X_{\tau_{i_1}(i_1 + 1)}} \circ s_{X_{i_1},X_{i_1 + 1}} = s_{X_{i_1 + 1}, X_{i_1}} \circ s_{X_{i_1}, X_{i_1 + 1}} = 1$ and obersve that it generalizes nicely when such an unreduced instance of juxtaposition sits between two reduced words in $F$ (possibly empty).

Then $(\tau^{\pm}_{i_n},...,\tau^{\pm}_{i_1}) = g_1n_1g^{-1}_1...g_mn_mg^{-1}_m$ where $f(n_i) = 0$ (this is what Mac Lane proves) implies that $s_{X_{\tau_{i_1}\left(...\left(\tau_{i_{n-1}}(i_n)\right)\right)}, X_{\tau_{i_1}\left(...\left(\tau_{i_{n-1}}(i_n + 1)\right)\right)}}\circ...\circ s_{X_{\tau_{i_1}(i_2)}, X_{\tau_{i_1}(i_2 + 1)}}s_{X_{i_1},X_{i_1 + 1}} = 0$.

I hope I didn't miss anything, but wouldn't by surprised if I did since it seems quite easy. Please, tell me if I'm wrong.

The key difficulty seems that $M$ is not a group, so we need to work with relevant properties of would-be group homomorphism $f\colon F\to M$ by hand.

  • $\begingroup$ You also need $g_i n_i^{\pm 1} g_i^{-1}$, right? $\endgroup$ – Martin Brandenburg Jan 13 '20 at 13:10
  • $\begingroup$ I don't understand the proof from "so it's possible that it doesn't translate to a product of conjugates ... " onwards. $\endgroup$ – Martin Brandenburg Jan 13 '20 at 13:11
  • $\begingroup$ @MartinBrandenburg Yeah, you need $n^{\pm}$, but I don't think that it causes probelms here. As for your misiunderstanding, it's better to give an example. Suppose that the reduced word $(\tau^{\pm}_{i_n},...,\tau^{\pm})$ equals the conjugate product $(\tau^{\pm}_{j_m},...,\tau^{\pm}_{j_1})(\tau_i)(\tau_{i+1})(\tau_i)(\tau_{i + 1})(\tau_i)(\tau_{i+1})((\tau^{\pm}_{j_1},...,\tau^{\pm}_{j_m})$ where $(\tau^{\pm}_{i_n},...,\tau^{\pm})$ is reduced. Since $\tau_i \neq \tau_{i + 1}$ in $F$, $(\tau_i)(\tau_{i+1})(\tau_i)(\tau_{i + 1})(\tau_i)(\tau_{i+1}) = $(contd) $\endgroup$ – Jxt921 Jan 13 '20 at 13:24
  • $\begingroup$ @MartinBrandenburg $= (\tau_i,\tau_{i+1},\tau_i,\tau_{i + 1},\tau_i,\tau_{i+1})$ is also reduced, thus the product is $(\tau^{\pm}_{j_m},...,\tau^{\pm}_{j_1})(\tau_i,\tau_{i + 1},\tau_i,\tau_{i + 1},\tau_i,\tau_{i+1})(\tau^{\pm}_{j_1}...\tau^{\pm}_{j_m})$. But it is possible that $\tau^{\pm}_{j_1} = \tau^{-1}_i$ or $\tau^{\pm}_{j_m} = \tau_i$. $\endgroup$ – Jxt921 Jan 13 '20 at 13:27
  • $\begingroup$ @MartinBrandenburg What I showed in the example with $f(\tau_i,\tau^{-1}_i)$ shows that this causes no problems since $f(\tau_i,\tau^{-1}_i) = f() = 1_{X_1\otimes ... \otimes X_n} = s_{X_{i_1 + 1}, X_{i_1}} \circ s_{X_{i_1}, X_{i_1 + 1}} = s_{X_{\tau_{i_1}(i_1)}, X_{\tau_{i_1}(i_1 + 1)}} \circ s_{X_{i_1},X_{i_1 + 1}}$ which is the value of $(\tau_i,\tau^{-1}_i)$ under $f$ as if $(\tau_i,\tau^{-1}_i)$ were reduced. I checked on the paper that it trivially generalizes to the case where such "unreduced" instances in $g_1n^{\pm}g^{-1}_1...g_mn^{\pm}g^{-1}_m$ apper between the reduced ones. $\endgroup$ – Jxt921 Jan 13 '20 at 13:30

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