An isomorphism in a note by Witherspoon

Question

Let $A$ be a Hopf algebra over a field $k$. In these notes by Witherspoon, we find the following lemma

For the second part, the hint given is that $A$-intertwiners from $U$ to $V$ are the same as linear maps $f$ satisfying $a . f = \varepsilon(a) f$, where $(a . f)(u) = a_{(1)} f(S(a_{(2)}) u)$.

Where do we need the bijectivity of the antipode? I don't think we do, and here is my computation.

Let $\phi$ and $\psi$ be the maps from left to right and right to left in the upper displayed equation, i.e. $\phi(f)(u)(v) = f(u \otimes v)$ and $\psi(g)(u \otimes v) = g(u)(v)$. Let $f \colon U \otimes V \to W$ be an $A$-module map, i.e. $a.f = \varepsilon(a) f$. Then \begin{align*} (a . \phi(f))(u)(v) &= [a_{(1)} . \phi(f)(S(a_{(2)}) u) ](v) \\ &= a_{(1,1)} \phi(f)(S(a_{(2)}) u)( S(a_{(1,2)}) v) \\ &= a_{(1)} f(S(a_{(3)}) u \otimes S(a_{(2)}) v) \\ &= a_{(1)} f(S(a_{(2)})_{(1)} u \otimes S(a_{(2)})_{(2)} v) \\ &= a_{(1)} f(S(a_{(2)}) (u \otimes v)) \\ &= a_{(1)} S(a_{(2)}) f(u \otimes v)) \\ &= \varepsilon(a) f(u \otimes v)) \\ &= \varepsilon(a) \phi(f)(u)(v)\ , \end{align*} so no bijectivity needed here. Assume then that $g \colon U \to \operatorname{Hom}_k(V, W)$ is $A$-linear. Then \begin{align*} (a . \psi(g))(u \otimes v) &= a_{(1)} \psi(g)(S(a_{(2)}) (u \otimes v)) \\ &= a_{(1)} \psi(g)(S(a_{(2)})_{(1)} u \otimes S(a_{(2)})_{(2)} v)) \\ &= a_{(1)} \psi(g)(S(a_{(2,2)}) u \otimes S(a_{(2,1)}) v)) \\ &= a_{(1)} g(S(a_{(2,2)}) u)(S(a_{(2,1)}) v)) \\ &= a_{(1,1)} g(S(a_{(2)}) u)(S(a_{(1,2)}) v)) \\ &= [ a_{(1)} . g(S(a_{(2)}) u) ] (v)) \\ &= ( a . g)(u)(v) \\ &= \varepsilon(a) g(u)(v) \\ &= \varepsilon(a) \psi(g)(u \otimes v) \ , \end{align*} and evidently I didn't need bijectivity of $S$ here either.

What's up with this?

Answer 1

As far as I’m aware, we don’t need the the antipode to be bijective. I also looked into Radford’s Hopf Algebras, where I found the following:

Suppose that $H$ is any Hopf algebra with antipode $S$ and $M$ and $N$ are left $H$-modules. For $a ∈ H$ and $f ∈ \operatorname{Hom}(M, N)$ we define $a \bullet f ∈ \operatorname{Hom}(M, N)$ by $$ (a \bullet f)(m) = a_{(1)} ⋅ f(S(a_{(2)}) ⋅ m) $$ for all $m ∈ M$. The proof of the following lemma is straightforward and left to the reader.

Lemma 10.3.1. Suppose that $H$ is a Hopf algebra with antipode $S$ over the field $k$. Let $M$ and $N$ be left $H$-modules. Then:

1. $(\operatorname{Hom}(M, N), \; \bullet)$ is a left $H$-module.
2. $a ⋅ f(m) = (a_{(1)} \bullet f)(a_{(2)} ⋅ m)$ for all $a ∈ H$, $f ∈ \operatorname{Hom}(M, N)$, and $m ∈ M$.
3. $f ∈ \operatorname{Hom}(M, N)$ is a module map if and only if $a \bullet f = \epsilon(a) f$ for all $a ∈ H$.

So it seems that the bijectivity of the antipode is indeed not needed. I assume that Witherspoon just mixed something up when editing her notes.

PS: That intertwiners are the same as invariant maps is also discussed and proven in “For two modules over a Hopf algebra $H$, are the module homomorphisms the same as the $H$-invariant linear maps?”, without any further assumptions on the antipode.