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 bradings 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 see 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?

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?

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

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 bradings 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 see 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?

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 bradings 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 see 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?