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I am quite confused on page 22,23 of the unit axiom of a monoidal category.

Unit axiom: $L_1:X \mapsto 1 \otimes X$ and $R_1:X \mapsto X \otimes 1$ are autoequivalences of $C$.

(end of pg 22) Choose natural isomorphisms $l_1:L_1 \Rightarrow \mathrm{id}$ such that it equals the diagram of compositions $$1 \otimes (1\otimes X) \rightarrow (1 \otimes 1) \otimes X \rightarrow 1 \otimes X. $$

Why can we do this, and why is it a natural isomorphism?


This doesn't seem right. What I see here: given a functor $F:C \rightarrow D$, we may define $i:F \Rightarrow \mathrm{id}$, if there exists an isomorphism $i_X:F(X) \rightarrow X$ for each $X \in C$?

But this would not imply commutativity of the diagram. Since we require $i_Y F(f) = f \, i_X$ given a morphism $f:X \rightarrow Y$. So $F(f)$ would be determined, i.e. we cannot define a natural isomorphism from arbitrary $i_X$.

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  • $\begingroup$ "Auto-", meaning "self"; an autoequivalence is an equivalence of $C$ with itself. $\endgroup$ – Malice Vidrine Sep 5 '18 at 9:42
  • $\begingroup$ Sorry, so what does"self " equivalent functors of $C$ mean? $\endgroup$ – CL. Sep 5 '18 at 9:44
  • $\begingroup$ It means an equivalence from $\mathcal{C}$ to itself. $\endgroup$ – Tobias Kildetoft Sep 5 '18 at 9:45
  • $\begingroup$ Do you know the definitions of functors, natural transformations, natural isomorphisms and equivalences of categories? An autoequivalence is an equivalence $\mathcal C\to\mathcal C$, i.e., a functor $F\colon\mathcal C\to\mathcal C$ such that there is a functor $G\colon\mathcal C\to\mathcal C$ and natural isomorphisms $FG\Rightarrow\operatorname{id}_{\mathcal C}$ and $GF\Rightarrow\operatorname{id}_{\mathcal C}$. $\endgroup$ – Christoph Sep 5 '18 at 9:48
  • $\begingroup$ @Christoph, oh, i see, thanks! $\endgroup$ – CL. Sep 5 '18 at 10:00
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The functor $(1\otimes-)\colon\mathcal{C}\to\mathcal{C}$ is an autoequivalence, hence the functor $(1\otimes-)^{\mathcal{C}}\colon\mathcal{C}^{\mathcal{C}}\to\mathcal{C}^{\mathcal{C}}$ is also an autoequivalence, hence $(1\otimes-)^{\mathcal{C}}$ is fully faithful and conservative. Thus, for every natural isomorphism $\gamma\colon(1\otimes(1\otimes-))\to(1\otimes-)$ there exists the unique natural isomorphism $\delta\colon(1\otimes-)\to I_{\mathcal{C}}$, such that $(1\otimes-)^{\mathcal{C}}(\delta)=\gamma$. So in your task you shoud only check that your $\gamma=(\imath\otimes I_{\mathcal{C}})\circ\alpha^{-1}(1,1,-)$ is a natural isomorphism. It follows from the following facts: $\alpha$ is a natural isomorphism, $\imath$ is an isomorphism, (horizontal and vertical) composition of natural isomorphisms is a natural isomorphism.

Appendix.

Definition 1. Let $\mathcal{A}$ and $\mathcal{B}$ be categories. An equivalence $4$-tuple between $\mathcal{A}$ and $\mathcal{B}$ is a $4$-tuple $(T,T',u_T,\varepsilon_T)$, such that $T\colon\mathcal{A}\to\mathcal{B}$ and $T'\colon\mathcal{B}\to\mathcal{A}$ are functors, $u_T\colon I_{\mathcal{A}}\to T'\circ T$ and $\varepsilon_T\colon T\circ T'\to I_{\mathcal{B}}$ are natural isomorphisms. In this case we say that $(T,T',u_T,\varepsilon_T)$ is an equivalence $4$-tuple for $T$.

Proposition 1. A functor $T$ is an equivalence iff there exists an equivalence $4$-tuple for $T$.

Proposition 2. Let $\mathcal{A}$, $\mathcal{B}$, $\mathcal{C}$, $\mathcal{D}$ be categories, $T\colon\mathcal{A}\to\mathcal{C}$ and $S\colon\mathcal{D}\to\mathcal{B}$ be functors. Then if the functors $T$ and $S$ are equivalences, then the functor $T^S\colon\mathcal{A}^{\mathcal{B}}\to\mathcal{C}^{\mathcal{D}}$ is also an equivalence.

Proof. If $(T,T',u_T,\varepsilon_T)$ is an equivalence $4$-tuple for $T$ and $(S,S',u_S,\varepsilon_S)$ is an equivalence $4$-tuple for $S$, then $(T^S,T'^{S'},u_T^{\varepsilon_S^{-1}},\varepsilon_T^{u_S^{-1}})$ is an equivalence $4$-tuple for $T^S$.

Corollary 1. Let $\mathcal{A}$ be a category, $T\colon\mathcal{A}\to\mathcal{A}$ be an endofunctor on $\mathcal{A}$. Then if $T$ is an autoequivalence, then $T^{\mathcal{A}}\colon\mathcal{A}^{\mathcal{A}}\to\mathcal{A}^{\mathcal{A}}$ is also an autoequivalence.

Proof. By the Proposition 2 the functor $T^{\mathcal{A}}=T^{I_{\mathcal{A}}}$ is an equivalence.

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  • $\begingroup$ Sorry, may you elaborate a little on $(1 \otimes - )^C : C^C \rightarrow C^C$ is also an autoequivalence? $\endgroup$ – CL. Sep 7 '18 at 11:08
  • $\begingroup$ @CyrylL. see Appendix. $\endgroup$ – Oskar Sep 7 '18 at 12:06

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