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I asked this a week ago at MSE, but nobody replied. Could anybody enlighten me if the following is true?

Let $F$ and $G$ be Hopf algebras over a field $k$ (in the usual sence, i.e. Hopf algebras in the category of vector spaces over $k$). Let $$ \varphi:F\to G $$ be a morphism of Hopf algebras (i.e. $\varphi$ preserves multiplication, comultiplication, unit, counit and antipode), and $$ H\subseteq G $$ a Hopf subalgebra in $G$ (i.e. $H$ is a Hopf algebra with respect to the multiplication, comultiplication, unit, counit and antipode induced from $G$).

Question:

Is it true that the preimage of $H$, $$ \varphi^{-1}(H)=\{x\in F:\ \varphi(x)\in H\} $$ is a Hopf subalgebra in $F$?

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    $\begingroup$ I think you can get a counterexample by looking at a special case when $F$ and $G$ are group algebras, $\varphi$ is induced by a surjective homomorphism of groups and $H$ is subalgebra spanned by the unit element. $\endgroup$
    – Victor Ostrik
    Commented Aug 25, 2023 at 17:12
  • $\begingroup$ @VictorOstrik I would think that in this case $\varphi^{-1}(H)$ is the group algebra of the kernel of this homomorphism of groups. That is not true? $\endgroup$
    – Sergei Akbarov
    Commented Aug 25, 2023 at 17:41
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    $\begingroup$ Assume that two elements $g$ and $h$ of the first group map to the same non-unit element of the second group. Then $g-h$ is in the pre-image. $\endgroup$
    – Victor Ostrik
    Commented Aug 25, 2023 at 18:02
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    $\begingroup$ @VictorOstrik ah, yes, so this means that $\varphi^{-1}(H)$ is bigger than the group algebra of the kernel... But why isn't $\varphi^{-1}(H)$ necessarily a Hopf subalgebra? $\endgroup$
    – Sergei Akbarov
    Commented Aug 25, 2023 at 18:20
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    $\begingroup$ For example assume that first group is of order 4 and the second group is of order 2. Then dimension of $\varphi^{-1}(H)$ is 3. This can't be Hopf subalgebra -- e.g. by Nichols-Zoeller theorem dimension of any Hopf subalgebra divides the dimension of Hopf algebra. $\endgroup$
    – Victor Ostrik
    Commented Aug 25, 2023 at 21:34

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No, it is only a coideal subalgebra. The subalgebra bit is clear (I hope).

Let us see why it is a coideal. This is a general coalgebraic fact. I keep your letters but assume they are all coalgebras only. I pick a basis of $F$ consisting of three types of elements:

  • $a_i$ - a basis of the kernel $\ker(\varphi)$,
  • $b_j$ - extend it to a basis of $\varphi^{-1}(H)$,
  • $c_k$ - extend it to a basis of $F$. Now pick $x\in \varphi^{-1}(H)$. What do we know about $\Delta (x)$? We know that $\varphi\otimes\varphi (\Delta (x))= \Delta (\varphi (x)) \in H \otimes H$. This means $\Delta (x)$ is a linear combination of the following basis elements of $F\otimes F$: $$a_i\otimes c_j, \ a_i \otimes b_j, \ a_i \otimes c_j,\ b_i \otimes a_j,\ b_i \otimes b_j,\ c_i \otimes a_j.$$ All of them belong to $F\otimes \varphi^{-1}(H) + \varphi^{-1}(H)\otimes F$, so it is a coideal.

It remains to see why it is not a subcoalgebra. The example by Ostrik in the comments is perfect. Let $F=kF_1$ and $G=kG_1$ be group algebras. All its Hopf subalgebras are group subalgebras and the homomorphisms are $\varphi =k \phi$ for a group map $\phi$. Now pick a subgroup $H_1 \leq G_1$, then $$\varphi^{-1} (H)= k\phi^{-1}(H_1) + \mbox{Span} \{ x-xy \,\mid\, x \in F_1, y \in \ker(\phi)\}.$$ This is a Hopf subalgebra if and only if $\phi$ is injective or $H_1=G_1$.

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  • $\begingroup$ Excuse me, why in the last equality the embedding $\subseteq$ holds? I.e., why $\varphi^{-1}(H)\subseteq...$ ? $\endgroup$
    – Sergei Akbarov
    Commented Aug 31, 2023 at 13:49
  • $\begingroup$ Think of an element in $\varphi^{-1}(H)$. It is a linear combination of elements of $F_1$. Separate them into two summands: $x$ -- linear combination of elements of $\phi^{-1}(H_1)$ and the rest $y$. Then $x$ belongs to the first summand, while $y$ belongs to $\ker (\varphi)$, which is the second summand. $\endgroup$
    – Bugs Bunny
    Commented Aug 31, 2023 at 15:18

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