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Khovanov's Frobenius algebra (used in the definition of Khovanov homology) is $\mathbb{Z}[X]/X^2$ with the comultiplication. $\Delta(X)=X\otimes X, \Delta(1)=1\otimes X+X\otimes 1$ and the trace $\epsilon(1)=0, \epsilon(X)=1$. (Although it is a bit more general but for my question this suffices.) The definition of a "self dual Frobenius extension" is given in and it is basically an isomorphism of the algebra with its dual under which multiplication is sent to comultiplication and the unit to the trace. This is easily seen to be the case for the Khovanov algebra with the isomorphism that sends $1$ to $X$ and vice-verse.

Now my question is whether $\Delta$ is sent to the multiplication $m$ under this isomorphism. $\Delta^*$ sends $1\otimes X+X\otimes 1 \to X$, $1\otimes 1 \to 1$ and $X\otimes X \to 0$ but this is not exactly $m$.

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I presume that by "the isomorphism that sends $1$ to $X$ and vice versa", you mean the isomorphism that sends $1$ to $X^\vee$ and $X$ to $1^\vee$, where $\{1^\vee,X^\vee\}$ is the dual basis to $\{1,X\}$?

In that case, it's just a calculation to check that this interchanges multiplication and comultiplication, and unit and trace. In fact, since the isomorphism between ${\mathbb Z}[X]/X^2$ and its dual is given by a symmetric matrix, once you know that multiplication is sent to comultiplication, it follows automatically that comultiplication is sent to multiplication.

By the way, the definition in the paper you link to does require comultiplication to be sent to multiplication, and not just that multiplication is sent to comultiplication.

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