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. 


I have posted this question on Math Stack Exchange, but have not got any reply.The link is here <https://math.stackexchange.com/questions/2859327/quandle-homomorphism-does-not-always-induces-group-homomorphim-on-inner-automorp>.


Any suggestions or hints would be of great help. 



Quandle homomorphism does not always induces group homomorphism on inner automorphism groups of quandles
Let's make sure we agree on definitions.


A **quandle** is an algebraic structure $(A,\*)$ for which
each right multiplication map $S\_a(x)= x\*a$ by an element $a\in A$
is an automorphism fixing $a$. These right multiplications
are called **inner automorphisms** of the quandle, and **$\textit{Inn}(A)$**
is the group generated by these right multiplications.


Let $G$ be a nonabelian
group and define a quandle operation
on $G$ by $a\*b = b^{-1}ab$. Let $f\colon G\to G$
be a constant function whose range is $\{c\}$
for some noncentral element $c\in G$.
The function $f$ is a quandle endomorphism of $(G,\*)$, since quandles are idempotent.
Also $\widetilde{f}(S\_g) = S\_{f(g)}=S\_c$ for all $g\in G$,
even for $g=1$. So, if $g=1$ we have $\textrm{id}=S\_1=S\_g \neq S\_c$,
and $\tilde{f}(\textrm{id})=\tilde{f}(S\_g)=S\_{f(g)}=S\_c\neq \textrm{id}$.


