Showing the triple $(\hom(C,A),@,\mu \epsilon)$ defines an algebra 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".
 A: 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.
