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?
Thanks for your help.
 A: 
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 
$$\begin{CD}
1 \otimes 1 @>{\phi_{i}}>> 1' \otimes 1' \\
@ViVV @VVi'V \\
1 @>>\phi_i> 1'
\end{CD}
$$
commutes for each $i=1,2$, 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.
