Timeline for Mac Lane's proof of coherence for symmetric monoidal categories
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Jan 12, 2020 at 21:57 | comment | added | Martin Brandenburg | I meant $X = X_1 \sqcup X_2 \sqcup X_3$, sorry. For the rest, please try to think more about it. | |
| Jan 12, 2020 at 21:48 | comment | added | Jxt921 | Regarding uniqueness: my lack of experience shows, but I don't see at the moment what the existence of such morphism $X_1\otimes ... \otimes X_n\to X_{\sigma^{-1}(1)}\otimes ... \otimes X_{\sigma^{-1}(n)}$ has to do with the original statement of uniqueness of such morphism. | |
| Jan 12, 2020 at 21:42 | comment | added | Jxt921 | Sorry, but I don't quite get your example. First, where did $X_3$ come from (a typo?) If $X = X_1\sqcup X_2$, then $s_1$ maps $(X_1\sqcup X_2)\otimes (X_1\sqcup X_2)$ to itself, and I think $(X_1\sqcup X_2)\otimes (X_1\sqcup X_2) = (X_1,X_1)\sqcup(X_1,X_2)\sqcup (X_2,X_1)\sqcup (X_2,X_2)$. What am I missing? | |
| Jan 12, 2020 at 21:33 | comment | added | Martin Brandenburg | Regarding uniqueness: We can define $X_1 \otimes \cdots \otimes X_n \to X_{\sigma^{-1}(1)} \otimes \cdots \otimes X_{\sigma^{-1}(n)}$ as the unique morphism which makes the following diagram commutative: $$\begin{array}{ccc} X_1 \otimes \cdots \otimes X_n & \longrightarrow & X_{\sigma^{-1}(1)} \otimes \cdots \otimes X_{\sigma^{-1}(n)} \\ \downarrow && \downarrow \\ X^{\otimes n} & \xrightarrow{~\sigma_*~} & X^{\otimes n} \end{array}$$ So this is well-defined. | |
| Jan 12, 2020 at 21:27 | history | edited | Martin Brandenburg | CC BY-SA 4.0 |
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| Jan 12, 2020 at 21:25 | comment | added | Martin Brandenburg | 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. | |
| Jan 12, 2020 at 21:24 | comment | added | Martin Brandenburg | 1) Yes, 2) Yes, | |
| Jan 12, 2020 at 21:12 | comment | added | Jxt921 | 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 $(i,k_{j_i})$. | |
| Jan 12, 2020 at 20:57 | comment | added | Jxt921 | $X_1\otimes ... \otimes X_n \to X^{\otimes n}$ factors over $X_{\sigma(1)}\otimes ... \otimes X_{\sigma(n)}$, and why it would imply the uniqueness? | |
| Jan 12, 2020 at 20:56 | comment | added | Jxt921 | 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 | |
| Jan 12, 2020 at 20:25 | history | edited | Martin Brandenburg | CC BY-SA 4.0 |
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| Jan 12, 2020 at 20:18 | history | edited | Martin Brandenburg | CC BY-SA 4.0 |
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| Jan 12, 2020 at 20:12 | history | answered | Martin Brandenburg | CC BY-SA 4.0 |