Given Hopf $\mathbb{C}$-algebra $H$, it's Hopf dual $H^o$ is the largest Hopf algebra contained in $H^\*$, the $\mathbb{C}$-linear dual of $H$. (This is well known to be well-defined, see for example Sweedler book.) 


If $j:G \to H$ is a linear map, then we have dual linear map
$$
j^\*:H^\* \to G^\*, ~~~~~~~~~ f \mapsto f \circ j.
$$
If we restrict $j^\*$ to $H^o$, then does $j^\*(H^o)$ be contained in $G^o$?
If $j$ is algebra map, then it is easy to show that 
$$
\Delta(j^\*(f)) = j^\*(f\_{(1)}) \otimes j^\*(f\_{(2)}) \in G^o \otimes G^o,
$$
where we use Sweedler notation. However, it $j$ is not algebra map, then it is not clear what happens. My guess is that the dual is only "functorial" for algebra maps, but it is not clear for me.



Functoriality of the Hopf dual
-as suggested after the discussion in the comments-  

i am understanding that the OP is asking whether a linear map $j:G \to H$ is functorial, in the sense that the image of the dual map $j^\*:H^\* \to G^\*$ preserves the finite dual hopf algebra $H^\circ$, i.e. 
$
j^\*(H^\circ)\subseteq G^\circ
$


The answer is generally no -at least for an arbitrary linear map $j$- and in my understanding this has already been shown in the counterexample proposed by user66288 in the comments to the OP.


However, it will be the case, i.e. $j$ will be functorial if it vanishes on an ideal of finite codimension in $G$, since then the image of the finite dual $H^\circ$ will be inside the finite dual $G^\circ$, i.e. the image $j^\*(f)$ of an arbitrary element $f\in H^\circ$ will be vanishing on an ideal of finite codimension in $G$:
$
j^\*(f)=f\circ j\in G^\circ
$ 
for all $f\in H^\circ$.  

(Notice that in this situation we actually have that $j^\*(f)\in G^\circ$ 
for all $f\in H^\*$, and this is something which makes me wonder whether it is too restrictive, as i have already mentioned in the comments to the OP). 


On the other hand, this is not an IFF statement, in the sense of the first of my comments above:  

Consider an ideal $R$ of $G$ of finite codimension, not necessarily contained in the kernel of $j$. Then, those $f\in H^\circ$ for which $j(R)\subseteq ker f$ will be mapped in $G^\circ$ under $j^\*$. If this happens for all $f\in H^\circ$ then $j^\*(H^\circ)\subseteq G^\circ$, so we cannot necessarily conclude the converse statement:  

"*if $j$ is functorial then it vanishes on an ideal of finite codimension in $G$*" 


**Edit (Nov 17):** i was thinking that an iff characterization of the functoriality of $j$, is still possible, in the sense of the last paragraph above: 



> 
> a linear map $j:G\to H$, between two hopf algebras $H$, $G$ is functorial if and only if for any $f\in H^\circ$, there is an ideal $R\_f$ of finite codimension in $G$, which is mapped inside the kernel of $f$ under $j$, i.e. $j(R\_f)\in ker f$
> 
> 
> 


(I am not sure if this implies that $j$ should be an algebra map then). 


