Let $H$ be a complex of vector spaces over some field $k$ which is endowed with the structure of a Hopf algebra object. I have heard several times that if $H$ is concentrated in positive or negative degrees, then the cohomology in degree zero of $H$ is a Hopf algebra in the usual sense. How does one prove such a statement and why is the condition on the degrees important?

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    $\begingroup$ When restricted to these subcategories, the functor $H^0(-)$ is symmetric monoidal and so preserves the Hopf algebra structures. I'm really not sure that this is a research-level question though. $\endgroup$ – Denis Nardin May 6 at 11:30
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    $\begingroup$ Could you actually explain your answer? $\endgroup$ – hopfology May 6 at 16:28
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    $\begingroup$ (I've edited the previous comment in response to a flag.) Which part needs explaining? Do you know what a symmetric monoidal ($k$-linear) functor is? Do you understand that if a concept (such as the notion of Hopf algebra object) is definable in the language of symmetric monoidal $k$-linear categories, then a symmetric monoidal $k$-linear functor will map models of that concept in the domain category to ones in the codomain? $\endgroup$ – Todd Trimble Jun 8 at 11:26

If $C$ is a graded coalgebra (e.g. $C$= the homology of a d.g. Hopf algebra), then $C_0$ is not necesarily a subcoalgebra, because
$$\Delta(C_0)\subset (C\otimes C)_0=\oplus_{n\in\mathbb Z}C_n\otimes C_{-n}$$

For example, $H=k\{x,y\}/(x^2=y^2=xy+yx)$ is a graded Hopf algebra with $|x|=1$ and $|y|=-1$, both $x$ and $y$ primitives (in particular a d.g. Hopf algebra with $d=0$ and agree with its homology).

$H_0=k\oplus kx\wedge y$, and $$ \Delta(x\wedge y)= (x\otimes 1+1\otimes x)(y\otimes 1+1\otimes y)$$ $$= x\wedge y\otimes 1+1\otimes x\wedge y+ x\otimes y-y\otimes x \notin H_0\otimes H_0$$

Of course if $C=\oplus_{n\geq 0} C_n$, then $C_n=0$ for $n<0$ and $C_{-n}=0$ for $n>0$, so, the only nonzero summand in $\oplus_{n\in\mathbb Z}C_n\otimes C_{-n}$ is $C_0\otimes C_0$.


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