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

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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.

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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 $$\mu\circ m = M\circ (\mu\otimes \mu) = (m\otimes m)\circ (1\otimes \tau\otimes 1)\circ (\mu\otimes \mu).$$

Now do the same reasoning for the comultiplication, you’ll see that you get the same (it can almost be claimed from the obvious symmetry of the expression).

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:

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The proof consists of writing down the four commutative diagrams corresponding to third and fourth characterizations. You will see that commutativity of two of them implies commutativity if the other two and vice versa. Check out theorem 3.1.1 of the book Hopf algebras by Sweedler.