# How to proof equivalent condition of algebra morphism and coalgebra morphism about Hopf algebra

My question is why $\mu$ is multiplication preserving iff $\mu\circ m=(m\otimes m)\circ(1\otimes \tau \otimes1)\circ(\mu \otimes \mu)$ and this holds iff $m$ is comultiplication preserving, where $\tau$ is the switching morphism, $\tau(A\otimes B)=B \otimes A$. Any help will be appreciated.

In terms of elements, the multiplication on $B\otimes B$ is $(x\otimes y)\cdot (x'\otimes y') = (x\cdot x')\otimes(y\cdot y') = m(x,x')\otimes m(y,y')$, so in details the multiplication is $$(x\otimes y)\otimes (x'\otimes y')\mapsto x\otimes x'\otimes y\otimes y'\mapsto m(x,x')\otimes m(y,y'),$$ meaning it's $$B\otimes B\otimes B\otimes B\to B\otimes B\otimes B\otimes B\to B\otimes B$$ given by $M:= (m\otimes m)\circ (1\otimes \tau\otimes 1)$.
Now saying that $\mu$ is multiplication preserving means that $\mu(x\cdot y) = \mu(x)\cdot \mu(y)$, which is a shortcut for $\mu(m(x,y)) = M(\mu(x)\otimes \mu(y))$. This is exactly the condition $$\mu\circ m = M\circ (\mu\otimes \mu) = (m\otimes m)\circ (1\otimes \tau\otimes 1)\circ (\mu\otimes \mu).$$
The relation: $$\mu\circ m=(m\otimes m)\circ(Id\otimes \tau \otimes Id)\circ(\mu \otimes \mu)$$ is part of "compatibility" conditions between the algebraic and coalgebraic structures of $B$, in order for $B$ to be a bialgebra. Notice that both of the above maps (LHS and RHS) are of the form: $B\otimes B\rightarrow B\otimes B$. (see also the diagram below for a more detailed analysis of the maps). Let's compute the LHS: $$\mu\circ m(g\otimes h)=\mu(gh)=\sum (gh)_{(1)}\otimes (gh)_{(2)}$$ and the RHS: $$(m\otimes m)\circ(Id\otimes \tau \otimes Id)\circ(\mu \otimes \mu)(g\otimes h)= \\ (m\otimes m)\circ(Id\otimes \tau \otimes Id)\Big(\sum g_{(1)}\otimes g_{(2)}\otimes h_{(1)}\otimes h_{(2)}\Big)=(m\otimes m)\Big(\sum g_{(1)}\otimes h_{(1)}\otimes g_{(2)}\otimes h_{(2)}\Big)=\sum g_{(1)}h_{(1)}\otimes g_{(2)}h_{(2)}$$ so your condition (part of the definition of a bialgebra) can now be equivalently written, in terms of elements $g,h\in B$, as: $$\mu(gh)=\sum (gh)_{(1)}\otimes (gh)_{(2)}=\sum g_{(1)}h_{(1)}\otimes g_{(2)}h_{(2)}=\mu(g)\mu(h)$$ and actually says that: the comultiplication $\mu:B\rightarrow B\otimes B$ is an algebra homomorphism, between the algebras $B$ and $B\otimes B$ (equipped with its usual tensor product algebra structure), i.e.: $$\mu\circ m=\big[(m\otimes m)\circ(Id\otimes \tau \otimes Id)\big]\circ(\mu \otimes \mu)\Leftrightarrow \\ \\ \\ \\ \\ \\ \mu(gh)=\mu(g)\mu(h)$$ or equivalently: the multiplication $m:B\otimes B\rightarrow B$ is a coalgebra morphism between the coalgebras $B\otimes B$ and $B$, (where $B\otimes B$ is equipped with its usual tensor product coalgebra structure), i.e.: $$\mu\circ m=(m\otimes m)\circ\big[(Id\otimes \tau \otimes Id)\circ(\mu \otimes \mu)\big]\Leftrightarrow \\ \\ \\ \\ \\ \\ \sum (gh)_{(1)}\otimes (gh)_{(2)}=\sum g_{(1)}h_{(1)}\otimes g_{(2)}h_{(2)}$$ An equivalent formulation of the above, is the commutativity of the following diagram: