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Let $(C, \otimes, 1, \alpha, l, r)$ be a monoidal category. I would like to prove that

the unit object is unique up to isomorphism.

More precisely, I need to show that for any triples $(1, l, r)$ and $(1^{\prime}, l^{\prime}, r^{\prime})$, there is a unique isomorphism $\phi \colon 1^{\prime} \rightarrow 1$ such that for any $X \in \mathrm{Ob}(C)$ , $$ l_X^{\prime} = l_X(\phi \otimes \mathrm{id}_X) \quad\text{and}\quad r_X^{\prime} = r_X(\mathrm{id}_X \otimes \phi) \,. $$

I found a proof in the book “Tensor Categories” that confuses me. The authors write that using commutativity of the triangle diagrams one has to show that $\phi$ maps $1 \otimes 1 \rightarrow 1$ to $ 1^{\prime} \otimes 1^{\prime} \rightarrow 1^{\prime}$. To prove that $\phi$ is the unique isomorphism having this property, it suffices to show that if $b \colon 1 \rightarrow 1$ is an isomorphism so that $$ \require{AMScd} \begin{CD} 1 \otimes 1 @> b \otimes b >> 1 \otimes 1 \\ @V \iota VV @VV \iota V \\ 1 @>> b > 1 \end{CD} $$ commutes, then $b = \mathrm{id}_1$. Can someone explain me why this shows that $\phi$ is unique?

Thanks for your help.

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  • $\begingroup$ If you know that the given conditions imply $\phi = r_1' \circ l_{1'}^{-1}$, doesn't that automatically give the uniqueness part? $\endgroup$ Apr 17, 2018 at 23:43
  • $\begingroup$ @Daniel Schepler In fact, you are right. $\endgroup$
    – Crystal
    Apr 18, 2018 at 13:27

1 Answer 1

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Can someone explain me why this shows that $\phi$ is unique ?

If you look carefully the requirement that

$b \colon 1 \to 1$ is an isomorphism so that $$\require{AMScd}\begin{CD}1 \otimes 1 @>b \otimes b >> 1 \otimes 1 \\ @ViVV @VViV\\ 1 @>>b> 1\end{CD}$$ commutes

basically means that $b$ is an isomorphism of the internal monoid $(1,i)$.

Requiring that such a $b$ is equal to $\text{id}_1$ basically means that $(1,i)$ has a trivial group of automorphisms.

From this it easily follows the uniqueness part of your $\varphi$. Assume that $\phi_1,\phi_2 \colon 1 \to 1'$ are two isomorphisms such that the required equations holds, then in particular you would have that $$\begin{CD} 1 \otimes 1 @>{\phi_{j}}>> 1' \otimes 1' \\ @ViVV @VVi'V \\ 1 @>>\phi_j> 1' \end{CD} $$ commutes for each $j=1,2$, hence $\phi_1$ and $\phi_2$ would be two isomorphisms between the monoids $(1,i)$ and $(1',i')$, then $\phi_2^{-1} \circ \phi_1 \colon 1 \to 1$ would be an automorphism of $1$, but that it must be equal to $\text{id}_1$. From this and from the fact that $\phi_1$ and $\phi_2^{-1}$ are both isomorphisms it follows that they are one inverse to each other and so $\phi_1=\phi_2$.

Hope this helps.

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  • $\begingroup$ Many thanks ! Your great answer really helped me. I am very happy now :). $\endgroup$
    – Crystal
    Apr 18, 2018 at 14:03
  • $\begingroup$ @CrystalCr glad to be of help. $\endgroup$ Apr 18, 2018 at 18:46

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