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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.

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    $\begingroup$ If $j$ is not an algebra map, I do not see a reason why $j^*(H^o)\subseteq G^o$. Indeed, let, for example, be $j\colon k[\mathbb{Z}]\to k[\mathbb{Z}]$ given by the projection map onto $ke_0$. Let $\epsilon\in k[\mathbb{Z}]^o$ - the counit. Then, $\epsilon\circ j\notin k[\mathbb{Z}]^o$ since the only ideal contained in $\operatorname{ker}\epsilon\circ j$ is $0$ which has not finite codimension in $k[\mathbb{Z}]$. $\endgroup$
    – Christoph Mark
    Commented Nov 11, 2018 at 17:06
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    $\begingroup$ $k[\mathbb{Z}]$ has a basis $e_n$ where $n\in\mathbb{Z}$ which multiplies like $e_ne_m=e_{n+m}$. $e_0$ is the unit with respect to this multiplication. @KonstantinosKanakoglou $\endgroup$
    – Christoph Mark
    Commented Nov 14, 2018 at 11:31
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    $\begingroup$ Ok it is clear now. Thanks. I am not sure if the OP is looking for a counterexample but In my understanding, your comment would probably deserve to be an answer. $\endgroup$
    – Konstantinos Kanakoglou
    Commented Nov 14, 2018 at 13:49
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    $\begingroup$ Generaly, the situation described in the OP, happens for those elements $f\in H^\circ$ for which there exists an ideal of finite codimension in $G$, whose image under $j$ is contained in the kernel of $f$. For example for those linear maps $j$ which vanish on an ideal of finite codimension in $G$ this will always be the case thus $j^*(H^\circ)$ will be inside $G^\circ$. I am not sure however if this is too restrictive for an answer. $\endgroup$
    – Konstantinos Kanakoglou
    Commented Nov 14, 2018 at 14:02
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    $\begingroup$ Well, you already proved in your post that $j^*$ is functorial if $j$ is an algebra map (regardless wether $j$ vanishes on an ideal of finite codimension in $G$). But you can't make this an "iff"-statement because $j^*$ will always be functorial in the finite-dimensional case. If $j$ is not an algbera map, it may or may not happen that $j^*$ is functorial (see my example above in the comments). $\endgroup$
    – Christoph Mark
    Commented Nov 15, 2018 at 11:32

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-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).

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  • $\begingroup$ In the finite-dimensional case, the zero ideal trivially has finite codimension. $\endgroup$
    – lambda
    Commented Nov 16, 2018 at 3:59
  • $\begingroup$ you are right. so i reedited and removed the last comment. $\endgroup$
    – Konstantinos Kanakoglou
    Commented Nov 16, 2018 at 12:17

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