Showing the triple $(\hom(C,A),@,\mu \epsilon)$ defines an algebra
Let $(C,\Delta,\epsilon)$ be a colalgebra and $(A, \mu, \nu)$ be an algebra where $\Delta, \mu$ are the coproduct and product whilst $\epsilon, \nu$ are the counit and unit.
Define the convolution $@$ for $f,g \in \hom(C,A)$ by $$(f @ g)(x) = \mu (f \otimes g) \Delta(x)$$
Now, the triple $(\hom(C,A),@,\nu \epsilon)$ defines an algebra. A neccessary condition in showing this is showing that $\mu \epsilon$ is a left\right unit. To show that it is a left unit, observe that:
$$((\mu \epsilon) @ f)(x) = \Sigma_{(x)} \epsilon(x')f(x'')=f(\Sigma_{x}\epsilon(x')x'')=f(x)$$
Can somebody explain to me the first equality $$((\mu \epsilon) @ f)(x) = \Sigma_{(x)} \epsilon(x')f(x'')$$
Why can we simply drop the $\mu$? I feel like this SHOULD read as
$$((\mu \epsilon) @ f)(x) = \Sigma_{(x)} (\mu \epsilon)(x')f(x'').$$
But alas, it does not. This is on page 50 Proposition 3.1 in Christian Kassel's "Quantum Groups".
I will denote the convolution product on $\operatorname{hom}(C, A)$ by $*$ instead of $@$, because $@$ hurts my eyes.
The calculation is Kassel’s book is slightly different than the one you have written down. Kassel states that $$ ( (\eta \varepsilon) * f )(x) = \sum_{(x)} \varepsilon(x') f(x'') = f \left( \sum_{(x)} \varepsilon(x') x'' \right) = f(x) \,. $$ Note that we are using $\eta \varepsilon$, not $\mu \varepsilon$. The composite $\mu \varepsilon$ doesn’t even make sense because $\varepsilon$ goes from $C$ to $k$ (the ground field) while $\mu$ goes from $A \otimes A$ to $A$.
To understand the first equality in this calculation we need to use the formula for the convolution product. The convolution $(\eta \varepsilon) * f$ is defined as the composite $$ C \xrightarrow{\enspace \Delta \enspace} C \otimes C \xrightarrow{\enspace (\eta \varepsilon) \otimes f \enspace} A \otimes A \xrightarrow{ \enspace \mu \enspace } A \,. $$ We therefore have \begin{align*} ( (\eta \varepsilon) * f )(x) &= [ \mu \circ ( (\eta \varepsilon) \otimes f ) \circ \Delta ](x) \\ &= [ \mu \circ ( (\eta \varepsilon) \otimes f ) ]( \Delta(x) ) \\ &= [ \mu \circ ( (\eta \varepsilon) \otimes f ) ]\left( \sum_{(x)} x' \otimes x'' \right) \\ &= \mu\left( ((\eta \varepsilon) \otimes f)\left( \sum_{(x)} x' \otimes x'' \right) \right) \\ &= \mu\left( \sum_{(x)} (\eta \varepsilon)(x') \otimes f(x'') \right) \\ &= \sum_{(x)} (\eta \varepsilon)(x') f(x'') \\ &= \sum_{(x)} \eta(\varepsilon(x')) f(x'') \,. \end{align*} The map $\eta$ is the unit of the algebra $A$. This map is given by $$ \eta(\lambda) = \lambda 1_A $$ for all $\lambda \in k$, where the multiplication on the right hand side denotes the scalar multiplication on $A$. We therefore have $$ \eta(\lambda) y = (\lambda 1_A) y = \lambda (1_A y) = \lambda y $$ for all $\lambda \in k$ and all $y \in A$. This explans why we can “simply drop the $\eta$”. By inserting this in the above calculation we arrive at $$ ( (\eta \varepsilon) * f )(x) = \sum_{(x)} \eta(\varepsilon(x')) f(x'') = \sum_{(x)} \varepsilon(x') f(x'') \,. $$ This is precisely the identity that Kassel uses in the first step of his calculation.