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Stream: learning: questions

Topic: SMCs as opfibrations

Joe Moeller (Feb 23 2021 at 06:45):

I'm reading Construction 2.0.0.1 in Lurie's Higher Algebra:

Let $(C, \otimes)$ be a symmetric monoidal category. We define a new category $C^\otimes$ as follows:

(i) An object of $C^\otimes$ is a finite sequence of objects of $C$, which we will denote by $[C_1, \dots, C_n]$.

(ii) A morphism from $[C_1, \dots, C_n]$ to $[C'_1, \dots, C'_m]$ in $C^\otimes$ consists of a subset $S \subseteq \{1, \dots, n\}$, a map of finite sets $\alpha \colon S \to \{1, \dots, m\}$, and a collection of morphisms $\{f_j \colon \bigotimes_{\alpha(i) = j} C_i \to C'_j\}_{1 \leq j \leq m}$ in the category $C$.

(iii) Suppose we are given morphisms $f \colon [C_1, \dots, C_n] \to [C'_1, \dots, C'_m]$ and $g \colon [C'_1, \dots, C'_m] \to [C''_1, \dots, C''_\ell]$ in $C^\otimes$, determining subsets $S \subseteq \{1, \dots, n\}$ and $T \subseteq \{1, \dots, m\}$ together with maps $\alpha \colon S \to \{1, \dots, m\}$ and $\beta \colon T \to \{1, \dots, \ell\}$. The composite $g \circ f$ is given by the subset $U = \alpha^{-1} T \subseteq \{1, \dots, n\}$, the composite map $\beta \circ \alpha \colon U \to \{1, \dots, \ell\}$, and the maps
$\bigotimes_{\beta \circ \alpha (i) = k} C_i \simeq \bigotimes_{\beta(j) = k} \bigotimes_{\alpha(i) = j} C_i \to \bigotimes_{} C'_j \to C''_k$
for $1 \leq k \leq \ell$.

I'm confused by this composition rule though. Consider $\mathsf{Set}$ with its cartesian monoidal structure. We can consider a function $f \colon A \times B \to C$ to be a morphism $[A,B] \to [C,D]$ is $\mathsf{Set}^\times$, and also a function $g \colon C \times D \to E$ as one $[C,D] \to [E]$. But what is the composite $[A,B] \to [E]$? $g(f(a,b),?)$ doesn't make sense to me.

Joe Moeller (Feb 23 2021 at 06:46):

image.png

Amar Hadzihasanovic (Feb 23 2021 at 07:01):

I think you're right. Perhaps he forgot to state that $\alpha$ needs to be a surjective map?

Amar Hadzihasanovic (Feb 23 2021 at 07:02):

If $\alpha$ and $\beta$ are surjective, the composition rule checks out.

John Baez (Feb 23 2021 at 07:09):

I don't think $\alpha$ needs to be surjective! But we need to read Lurie's definition carefully when $\alpha$ is not surjective, and use the fact that the tensor product of an empty set of objects is the unit object.

Joe Moeller said:

Let $(C, \otimes)$ be a symmetric monoidal category. We define a new category $C^\otimes$ as follows:

(i) An object of $C^\otimes$ is a finite sequence of objects of $C$, which we will denote by $[C_1, \dots, C_n]$.

(ii) A morphism from $[C_1, \dots, C_n]$ to $[C'_1, \dots, C'_m]$ in $C^\otimes$ consists of a subset $S \subseteq \{1, \dots, n\}$, a map of finite sets $\alpha \colon S \to \{1, \dots, m\}$, and a collection of morphisms $\{f_j \colon \bigotimes_{\alpha(i) = j} C_i \to C'_j\}_{1 \leq j \leq m}$ in the category $C$.

[....]

Consider $\mathsf{Set}$ with its cartesian monoidal structure. We can consider a function $f \colon A \times B \to C$ to be a morphism $[A,B] \to [C,D]$ in $\mathsf{Set}^\times$ [...]

How is that a morphism in $\mathsf{Set}^\times$? What's your map of finite sets $\alpha : \{A,B\} \to \{C,D\}$?

I'll guess $\alpha$ maps both $A$ and $B$ to $C$. Then to get a morphism we need a function $f_1 \colon A \times B \to C$ and also a morphism $f_2 \colon 1 \to D$. Here $1$ is the product of the set of objects that map to $D$... which is the empty set.

I think this should solve your problem.

Amar Hadzihasanovic (Feb 23 2021 at 07:23):

Of course! I have fallen for an old trap.

Joshua Meyers (Feb 23 2021 at 09:27):

Interesting construction, what is it being used for?

Joe Moeller (Feb 23 2021 at 16:36):

The category constructed (along with an opfibration to $\mathsf{Fin}_*$ that's not to hard to see) carries enough data to recover the original monoidal category. This characterization is the "right" way to think about symmetric monoidal categories if you want to jump up to $(\infty, 1)$-symmetric monoidal categories.

Mike Shulman (Feb 23 2021 at 18:53):

Or, at least, a right way.

Joe Moeller (Feb 23 2021 at 20:09):

John Baez said:

How is that a morphism in $\mathsf{Set}^\times$? What's your map of finite sets $\alpha : \{A,B\} \to \{C,D\}$?

I'll guess $\alpha$ maps both $A$ and $B$ to $C$. Then to get a morphism we need a function $f_1 \colon A \times B \to C$ and also a morphism $f_2 \colon 1 \to D$. Here $1$ is the product of the set of objects that map to $D$... which is the empty set.

I think this should solve your problem.

Oh, you're right. I was completely glossing over this, thinking $D$ just gets basically nothing.