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I am currently reading Hatcher's book, trying to understand the proof of Hopf's classification Theorem on Hopf algebras that says the following:

Every Hopf algebra $A$ that is commutative and associative over a field $F$ of characteristic 0 such that each $A^n$ is finitely generated is isomorphic to the tensor product of an exterior algebra on odd dimensional generators and a polynomial algebra on even dimensional generators.

Hatcher then defines $A_n$ to be the subalgebra generated by $x_1,x_2,....x_n$ with $x_i$ $\in$ $A^{|x_i|}$ and $|x_i| \leq |x_{i+1}|$ where $|x_i|$ denotes the the degree of $x_i$. Then Hatcher claims there are natural surjections from $A_{n-1} \otimes F{[x_n]}$ $\rightarrow$ $A_n$ when $|x_n|$ is even $A_{n-1}$ $\otimes$ $\wedge_F[x_n]$ $\rightarrow$ $A_n$ when $|x_n|$ is odd. What are those natural surjections and why do they depend on $|x_n|$?

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  • $\begingroup$ Unless your Hopf algebras are assumed to be (graded) commutative, this is false. $\endgroup$ – Najib Idrissi Feb 29 '16 at 16:46
  • $\begingroup$ yeah I just edited $\endgroup$ – TheGeometer Feb 29 '16 at 16:52
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So there is a "natural" in the colloquial sense homomorphism that is fairly easy to argue is surjective. Basically, there's an obvious inclusion of $A_{n-1}$ into $A_n$ and a homomorphism from $F[\alpha]_{|x_n|}$ or $\wedge_F[\alpha]_{|x_n|}$ into $A_n$ by evaluating at $x_n$. I'm using the notation $F[\alpha]_n$ for the ring of polynomials with indeterminate $\alpha$ with grade $n$, similarly for $\wedge_F$. We stick these together then multiply: $A_{n-1}\otimes F[\alpha]_{|x_n|}\to A_n\otimes A_n \to A_n$. It's pretty easy to argue from there that every element of $A_n$ is of the form $xP(x_n)$ for some $x\in A_{n-1}$ and polynomial $P$, and further that $P$ is a linear polynomial when $|x_n|$ is odd. With this, it's easy to see that the homomorphism is surjective.

Here's a more fleshed out abstract nonsense approach. Let $\mathcal{A}(S)$ be the free graded-commutative algebra on the graded set $S$. (A graded set is just a collection of sets indexed by grade.) In particular $F[\alpha]_n$ and $\wedge_F[\alpha]_n$ are $\mathcal{A}(\{\alpha\}_n)$ where the subscript $n$ means that set is at grade $n$. Let $S_n = \{x_i\ |\ i \leq n\}$ where $x_i$ are as in the question and the assumptions from the question are in force. (Incidentally, the theorem is then that every Hopf algebra under the assumptions is isomorphic to $\mathcal{A}(S)$ for some grade-wise finite $S$.) $A_n$, being finitely generated, is a quotient of $\mathcal{A}(S_n)$ which, by construction, makes the unique graded set function $S_n \to A_n$ corresponding to the quotient an inclusion. Being a left adjoint, $\mathcal{A}$ preserves coproducts and, in the category of graded-commutative algebras, tensor products are coproducts, so $$\mathcal{A}(S_n) \cong \mathcal{A}(S_{n-1}+\{x_n\}_{|x_n|}) \cong \mathcal{A}(S_{n-1})\otimes\mathcal{A}(\{x_n\}_{|x_n|})$$

Call the quotient maps $q : \mathcal{A}(S_{n-1})\twoheadrightarrow A_{n-1}$ and $q' : \mathcal{A}(S_n)\twoheadrightarrow A_n$. Coming from a coproduct, $q'$ is $[f,e] : \mathcal{A}(S_{n-1})\otimes\mathcal{A}(\{x_n\}_{|x_n|})\to A_n$ where $f : \mathcal{A}(S_{n-1})\to A_n$ and $e : \mathcal{A}(\{x_n\}_{|x_n|})\to A_n$. So the following diagram will commute

$$\require{AMScd} \begin{CD} \mathcal{A}(S_n) @= \mathcal{A}(S_n) \\ @Vq\otimes idVV @VVq'=[f,e]V \\ A_{n-1}\otimes\mathcal{A}(\{x_n\}_{|x_n|}) @>>[\iota,e]> A_n \end{CD}$$ where $\iota$ is the inclusion of $A_{n-1}$ into $A_n$, if this diagram does $$\begin{CD} \mathcal{A}(S_{n-1}) @= \mathcal{A}(S_{n-1}) \\ @VqVV @VVfV \\ A_{n-1} @>>\iota> A_n \end{CD}$$ in which case $[\iota,e]$ will be a surjection because $q'=[f,e]$ is. Both paths correspond to the same graded set function $S_{n-1}\to A_n$ by construction and so are equal by the universal property of $\mathcal{A}$, and thus both diagrams commute. Incidentally, $e$ is the evaluation map mentioned in the first part. $\iota$ is unique by the universal property of $q$ as a quotient, and $e$ is essentially given so the homomorphism $A_{n-1}\otimes\mathcal{A}(S_{n-1})\to A_n$ is unique. We can get the factorization via $A_n\otimes A_n$ in the first paragraph from the defining equation of $[\iota,e]$. $[\iota,e] = \varepsilon\circ(\iota\otimes e)$ where $\varepsilon = [id,id]$ is the counit of the adjunction defining coproducts and also corresponds to the multiplication homomorphism.

(It seems I lost the thread in this. The reason I did the abstract nonsense approach is that the phrase "natural surjection" often means the quotient map induced by a universal property, e.g. a coequalizer. In this case, the map in the question (which I called $[\iota,e]$) is not that but is closely related to such a map (namely $q'$)... until you prove the theorem, then it becomes clear that $[\iota,e]$ is just $q'$ from an isomorphic representation of $\mathcal{A}(S_n)$ and so is such a "natural surjection".)

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