Functoriality of the Hopf dual 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.
 A: -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).  
