$f:C\longrightarrow D$ coalgebra homomorphism $\Rightarrow$ $f^*:D^*\longrightarrow C^*$ algebra homomorphism I'm trying to prove the next problem using only arrows (avoiding Sweedler's notation):

If $f:C\longrightarrow D$ is a coalgebra homomorphism, then $f^*:D^*\longrightarrow C^*$ is an algebra homomorphism.

So here are the hipotesis and other definition arrows:
$C$ coalgebra $\Rightarrow$ $\exists$ $\Delta_C:C\longrightarrow C\otimes C$, $\epsilon_C:C\longrightarrow K$ such that: $(1_C\otimes \Delta_C)\Delta_C=(\Delta_C \otimes 1_C)\Delta_C$ and $(1_C\otimes \epsilon_C)\Delta_C=(\epsilon_C \otimes 1_C)\Delta_C$.
$D$ coalgebra $\Rightarrow$ $\exists$ $\Delta_D:D\longrightarrow D\otimes D$, $\epsilon_D:D\longrightarrow K$ such that: $(1_D\otimes \Delta_D)\Delta_D=(\Delta_D \otimes 1_D)\Delta_D$ and $(1_D\otimes \epsilon_D)\Delta_D=(\epsilon_D \otimes 1_D)\Delta_D$.
$f$ coalgebra homomorphism $\Rightarrow$ $\Delta_Df=(f\otimes f)\Delta_C$ and $\epsilon_Df=\epsilon_C$.
$D^*$ algebra $\Rightarrow$ $\exists$ $\mu_{D^*}:D^*\otimes D^*\hookrightarrow (D\otimes D)^*\xrightarrow{\Delta^*_D} D^*$, $\eta_{D^*}:K\cong K^*\xrightarrow{\epsilon^*_D} D^*$ such that $\mu_{D^*}(1_{D^*}\otimes \mu_{D^*})=\mu_{D^*}(\mu_{D^*}\otimes 1_{D^*})$ and $\mu_{D^*}(\eta_{D^*}\otimes 1_{D^*})=\mu_{D^*}(1_{D^*}\otimes \eta_{D^*})$.
$C^*$ algebra $\Rightarrow$ $\exists$ $\mu_{C^*}:C^*\otimes C^*\hookrightarrow (C\otimes C)^*\xrightarrow{\Delta^*_C} C^*$, $\eta_{C^*}:K\cong K^*\xrightarrow{\epsilon^*_C} C^*$ such that $\mu_{C^*}(1_{C^*}\otimes \mu_{C^*})=\mu_{C^*}(\mu_{C^*}\otimes 1_{C^*})$ and $\mu_{C^*}(\eta_{C^*}\otimes 1_{C^*})=\mu_{C^*}(1_{C^*}\otimes \eta_{C^*})$.
To show: $f^*$ is an algebra homomorphism, i.e., satisfies $\mu_{C^*}(f^*\otimes f^*)=f^*\mu_{D^*}$ and $f^*\eta_{D^*}=\eta_{C^*}$.
Can't seem to build the bridge between arrows, I need some hints on this. Thanks.
 A: Using Sweedler notation one can quickly check the commutativity of those diagrams involving the induced morphism of dual algebras. The OP asks for an answer without it, though.
Let us show that
$$\Delta^*_C\circ i_C\circ (f^*\otimes f^*)=f^*\circ \Delta^*_D\circ i_D,$$
where $i_C: C^*\otimes C^*\rightarrow(C\otimes C)^*$ and $i_D:D^*\otimes D^*\rightarrow(D\otimes D)^*$,
by dualizing  $$(f\otimes f)\circ \Delta_C=\Delta_D\circ f. $$
We already know that $\mu^*_C:=\Delta^*_C\circ i_C$ and similarly for $\mu^*_D$.
Dualization leads to
$$\Delta^*_C\circ (f\otimes f)^*=f^*\circ \Delta^*_D.$$
Then
$$\Delta^*_C\circ (f\otimes f)^*\circ i_D=f^*\circ \Delta^*_D\circ i_D. $$
The thesis follows by using
$$(f\otimes f)^*\circ i_D=i_C\circ (f^*\otimes f^*), $$
which follows from the very definition of the maps involved (and without using the Sweedler notation as no comultiplication is present). In fact, for all $\varphi_1\otimes\varphi_2\in D^*\otimes D^*$ we have
$$(f\otimes f)^*(i_D(\varphi_1\otimes\varphi_2))\in (C\otimes C)^*; $$
applying this map to any $c_1\otimes c_2\in C\otimes C$ we arrive at
$$(f\otimes f)^*(i_D(\varphi_1\otimes\varphi_2))(c_1\otimes c_2)=
\text{(definition of dual map)}=(i_D(\varphi_1\otimes\varphi_2))(f(c_1)\otimes f(c_2))=
\text{(definition of }~i_D)= \varphi_1(f(c_1))\varphi_2(f(c_2)).$$
Similarly,
$$i_C((f^*\otimes f^*)(\varphi_1\otimes\varphi_2))=i_C((f^*(\varphi_1)\otimes f^*(\varphi_2) )\in (C\otimes C)^*; $$
on any $c_1\otimes c_2\in C\otimes C$ it reads
$$i_C((f^*(\varphi_1)\otimes f^*(\varphi_2) )(c_1\otimes c_2)=\text{(definition of }~i_C)=
f^*(\varphi_1)(c_1)f^*(\varphi_2)(c_2)=\text{(definition of dual maps)}=
\varphi_1(f(c_1))\varphi_2(f(c_2)).$$
The diagram involving the dualized counits is treated similarly.
