Binary Erasure Channel Capacity From Cover&Thomas Information Theory Book 7.1.5,
Binary erasure channel erases $\alpha$ fraction of bits.
To calculate the capacity of the binary erasure channel, the text says:

Letting $E$ be the event $\{Y=e\}$, using the expansion
  $H(Y)=H(Y,E)=H(E)+H(Y|E)$

The second equality is the chain rule of joint entropy, so $H(Y,E)$ is joint entropy. Then where does $H(Y)=H(Y,E)$ come from?
And the text continues:

and letting $\text{Pr}(X=1)=\pi$, we have $H(Y)=H((1-\pi)(1-\alpha),\alpha,\pi(1-\alpha))=H(\alpha)+(1-\alpha)H(\pi)$.

I see that for the binary erasure channel, 0->0, {0,1}->e, 1->1 each has the probability of $(1-\pi)(1-\alpha), \alpha, \pi(1-\alpha)$, respectively. How can this lead to $H(\alpha)+(1-\alpha)H(\pi)$? And where is the former expansion used? 
 A: 
Letting $E$ be the event ${=}$, using the expansion
  $()=(,)=()+(|)$

To make the argument clearer, define the event $E$ as being $1$ if $Y = e$ and $0$ if $Y \neq e$. i.e. $E = f(Y)$ where
$$f(x) = \begin{cases} 
      1 & x = e \\
      0 & x \neq e      
   \end{cases} 
$$
From this it is clear that $H(E | Y)$ will be $0$ because $E$ is a function of $Y$ (i.e $E = f(Y)$). You can confirm this by explicitly computing $H(E | Y) = \sum _{i \in \{0, 1, e\}} p(Y = i) H(E | Y = i)$ if you want to.
Now lets look at the joint entropy $H(Y, E)$:
$H(Y, E) = H(Y) + H(E | Y) = H(Y)$
$H(Y, E) = H(E) + H(Y | E)$
$\implies H(Y) = H(E) + H(Y|E)$
The input distribution is defined as being $\text{Bern}(\pi)$ and so $p(E = 1) = \alpha\pi + (1 - \pi)\alpha = \alpha$ and $p(E = 0) = 1 - p(E = 1) = (1 - \alpha)$. 
It then follows that $H(E) = - \alpha\log \alpha - (1 - \alpha)\log (1 - \alpha) = H(\alpha)$. 
We now move on to computing $H(Y | E)$.
\begin{align}
H(Y|E) &= \textstyle \sum _{i \in \{0, 1\}} p(E = i) H(Y | E = i) \\
&= p(E = 0) H(Y | E = 0) + p(E = 1) H(Y | E = 1) \\
&= (1 - \alpha) H(Y | E = 0) + \alpha H(Y | E = 1) \\
&= (1 - \alpha) \left \{ \textstyle \sum_{y \in \mathcal{Y}} p(Y = y | E = 0) \log p(Y = y | E = 0) \right \} + \alpha \left \{ \textstyle \sum_{y \in \mathcal{Y}} p(Y = y | E = 1) \log p(Y = y | E = 1) \right \} \\
&= (1 - \alpha) \{ - \pi \log \pi - (1 - \pi)\log (1 - \pi)  - 0\log0\} + (\alpha \times 0) \\
&= (1 - \alpha)H(\pi)
\end{align}
From the above $H(Y) = H(E) + H(Y|E) = H(\alpha) + (1 - \alpha)H(\pi)$
