a {\ensuremath{\otimes}}b\,|\,({\ensuremath{\mathrm{id}_{}}}{\ensuremath{\otimes}}\delta^\dag) {\circ}(\delta {\ensuremath{\otimes}}{\ensuremath{\mathrm{id}_{}}})(a {\ensuremath{\otimes}}a) \rangle}} \\
    & = {\ensuremath{\langle ({\ensuremath{\mathrm{id}_{}}}{\ensuremath{\otimes}}\delta)(a {\ensuremath{\otimes}}b)\,|\,(\delta {\ensuremath{\otimes}}{\ensuremath{\mathrm{id}_{}}})(a {\ensuremath{\otimes}}a) \rangle}} \\
    & = {\ensuremath{\langle a {\ensuremath{\otimes}}b {\ensuremath{\otimes}}b\,|\,a {\ensuremath{\otimes}}a {\ensuremath{\otimes}}a \rangle}} \\
    & = {\ensuremath{\langle a\,|\,a \rangle}} \cdot {\ensuremath{\langle b\,|\,a \rangle}} \cdot {\ensuremath{\langle b\,|\,a \rangle}}.
  \end{aligned}$$ Analogously ${\ensuremath{\langle b\,|\,b \rangle}}{\ensuremath{\langle b\,|\,b \rangle}}{\ensuremath{\langle a\,|\,b \rangle}} =
  {\ensuremath{\langle b\,|\,b \rangle}}{\ensuremath{\langle a\,|\,b \rangle}}{\ensuremath{\langle a\,|\,b \rangle}}$. Hence, if ${\ensuremath{\langle a\,|\,a \rangle}}$ and ${\ensuremath{\langle b\,|\,a \rangle}}$ are both nonzero, then ${\ensuremath{\langle a\,|\,a \rangle}}={\ensuremath{\langle b\,|\,a \rangle}}$ and ${\ensuremath{\langle b\,|\,b \rangle}}={\ensuremath{\langle a\,|\,b \rangle}}$. So ${\ensuremath{\langle a\,|\,a \rangle}}, {\ensuremath{\langle b\,|\,b \rangle}} \in {\ensuremath{\mathbb{R}}}$ and ${\ensuremath{\langle a\,|\,a \rangle}}=
  {\ensuremath{\langle a\,|\,b \rangle}} = {\ensuremath{\langle b\,|\,a \rangle}} = {\ensuremath{\langle b\,|\,b \rangle}}$. Now suppose ${\ensuremath{\langle a\,|\,b \rangle}}\neq 0$. Then we can conclude ${\ensuremath{\langle a-b\,|\,a-b \rangle}} = {\ensuremath{\langle a\,|\,a \rangle}} - {\ensuremath{\langle a\,|\,b \rangle}} - {\ensuremath{\langle b\,|\,a \rangle}} +
  {\ensuremath{\langle b\,|\,b \rangle}} = 0$. So $a-b=0$, [*i.e.* ]{}$a=b$. Hence the copyables are pairwise orthogonal.

Applying dagger, a copyable element of $A$ corresponds exactly to a comonoid homomorphism $({\ensuremath{\mathbb{C}}}, \delta_{{\ensuremath{\mathbb{C}}}})
\to (A, \delta)$: $$\xymatrix{
  1 \ar@{|->}[d] \ar@{|->}[r] & 1 {\ensuremath{\otimes}}1 \ar@{|->}[d] \\  
  a \ar@{|->}[r] & a {\ensuremath{\otimes}}a.
}$$ We have already seen that copyable elements correspond exactly to algebra homomorphisms $$(A, \mu) \to ({\ensuremath{\mathbb{C}}}, \mu_{{\ensuremath{\mathbb{C}}}}),$$ [*i.e.* ]{}to *characters* of the 