Show that the unit object in a monoidal category is unique up to isomorphism.

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?

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

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.

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