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Let $X$ and $Y$ be two quandles and $f: X \rightarrow Y$ be a quandle homomorphism. Then we can define a map $\bar f: Inn(X) \rightarrow Inn(Y)$ as $\bar f(S_a)=S_{f(a)}$, where $a \in X$. Then $\bar f$ may not be a group homomorphism. But I am not able to construct such example. Can someone help me?

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There are many possible obstructions to what you propose, but the first obstacle is that your $\bar{f}$ may not be well defined. Remember that Inn$(X)$ is a group of automorphisms of $X$, and two different elements of $X$ may induce the same automorphism of $X$. In other words, $S_a$ and $S_b$ may be the same element of Inn$(X)$. For your $\bar{f}$ to be well-defined, you would have to show that the result does not depend on whether you choose $a$ or $b$ to represent that element of Inn$(X)$. But in fact, it might.

To see an example of this, start with the smallest quandle in which there are two distinct elements $a$ and $b$ such that $S_a=S_b,$ namely the only two element quandle, which I will call $2I$: you can call its elements $\{1,2\}$ and then it has the boring multiplication table $\begin{array}{cc}1 & 2 \\ 1&2 \end{array}$. Evidently $S_1 = S_2 = \,$Id. Now we just need to find another quandle $Q$ with a subquandle isomorphic to $2I$, but where those two elements act on the rest of $Q$ differently. The smallest example of this turns out to have five elements $\{1,2,3,4,5\}$ with multiplication table $$\begin{array}{ccccc} 1 & 2 & 4 & 5 & 3 \\ 1 & 2 & 5 & 3 & 4 \\ 2 & 1 & 3 & 5 & 4 \\ 2 & 1 & 5 & 4 & 3 \\ 2 & 1 & 4 & 3 & 5 \end{array}$$ The inclusion map $\iota:2I \rightarrow Q$ is of course a homomorphism, but now there is no way to tell whether $\bar{\iota}$ should map the identity in Inn$(2I)$ to the permutation $S_1 = (3 4 5)$ or $S_2=(3 5 4)$ in Inn$(Q)$.

(In case you are wondering where $Q$ came from, it is a sort of gluing together of $2I$ and the only latin quandle of order 3, called the dihedral quandle $D_3$ of order 3, which has a large enough automorphism group so that in the "sum," the two elements of the $2I$ component can act differently on the $D_3$ component.)

Even if $\bar{f}$ happens to be well-defined, there are other obstacles to it being a group homomorphism, but hopefully this example is illuminating.

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  • $\begingroup$ I should add, in case it's not clear, that in the case that $f$ is an epimorphism, $\bar{f}$ is indeed a well-defined group homomorphism. For if $S_a=S_b$, let $y=f(x)$ be an arbitrary element of $Y$. Then $f(a)\rhd y = f(a)\rhd f(x) = f(a\rhd x) = f(b\rhd x) = f(b)\rhd f(x) = f(b)\rhd y$, or in other words, $S_{f(a)} = S_{f(b)}$ in Inn$(Y)$. A similar calculation shows that if $S_c = S_bS_a$ in Inn$(X)$, then $S_{f(c)} = S_{f(b)}S_{f(a)}$ in Inn$(Y)$. Hence $\bar{f}$ is well defined on the generators of Inn$(X)$ compatibly with multiplication, and so extends to a (unique) group homomorphism. $\endgroup$ Jul 26, 2018 at 10:45

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