# 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.

• 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 $k[\mathbb{Z}]$. – user66288 Nov 11 '18 at 17:06
• $k[\mathbb{Z}]$ 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 – user66288 Nov 14 '18 at 11:31
• 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. – Konstantinos Kanakoglou Nov 14 '18 at 13:49
• 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. – Konstantinos Kanakoglou Nov 14 '18 at 14:02
• 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). – user66288 Nov 15 '18 at 11:32

-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).

• In the finite-dimensional case, the zero ideal trivially has finite codimension. – lambda Nov 16 '18 at 3:59
• you are right. so i reedited and removed the last comment. – Konstantinos Kanakoglou Nov 16 '18 at 12:17