Is the definition of the canonical projection given wrong ? If so, is there any problem in my correction? In the book of linear algebra by Werner Greub, at page $59$ it is given that

and from this definition, he argues that

However, by the definition of $\pi_\varrho$, whatever the map $x \in \oplus_\alpha E_\alpha$ is, it will map it to $x_\varrho$, so in $2.25$ if $\sigma$ is different from $\varrho$, then the composition will take any $x_\sigma$ any map it to $x_\varrho$, which indicates $2.25$ is wrong, so from this argument, I thought the definition of $\pi_\varrho$ is not complete and tried to redefine it by
$$\pi_\varrho (x_\alpha) = \delta_\alpha^\varrho * x_\varrho.$$
So the my question is that is my argument correct ? and also is my correction correct ? If not, what is the problem.
Note: I normally don't like to post a question with pictures, but while studying and trying to find an answer to some problems, it is really annoying to type everything every time if you are asking such questions regularly, so I'm hoping that it will not cause any problem.
Edit:
I think what I don't understand is that how is $x_\varrho$ in $2.24$ is determined.
 A: Consider what $\pi_\rho \circ i_\sigma$ is: it maps a point in $x \in E_\sigma$ to the function $i_\sigma(x)$ from $A \to \cup_\alpha E_\alpha$ that is $0$ for all $\alpha \neq \sigma$, so $i_\sigma(x)(\alpha) =0$ for $\alpha \neq \sigma$ and $i_\sigma(x)(\sigma) = x$.
Some motivation: $\oplus_\alpha E_\alpha$ can be seen as a formal sum of the vector spaces $E_\alpha$ : pick finitely many $\alpha_1, \ldots, \alpha_n$ from $A$ and a point $x_{\alpha_i} \in E_{\alpha_i}$ for $i=1,\ldots, n$ and we consider all such elements $x_{\alpha_1} + \ldots + x_{\alpha_n}$ as elements of $\oplus_\alpha E_\alpha$, but note that the sum is meaningless a priori. One $E_\alpha$ could be $\mathbb{R}^3$ and another $\mathbb{R}^4$ etc. We cannot really add them in a sum like that. So we consider these as "formal" sums, that we just compute with. To multiply a sum by a scalar, just do this for every summand, and we can add to finite formal sums just by collecting terms that happen to be in the same $E_\alpha$ (where thay can be added properly), and leaving terms from different ones as they were. The trick with defining it as the text is a constructive way to make this more exact: the above formal sum $x_{\alpha_1} + \ldots + x_{\alpha_n}$ corresponds to the function $x$ that sends $\alpha_i$ to $x_{\alpha_i}$ for all finitely many $i$ and to the $0$ of $E_\alpha$ for all other $\alpha$. We only take finite sums, hence condition (ii). We can think of any element as an infinite sum, in a way, namely we pick $0$ for all $\alpha$ that are not in our finite sum $x_{\alpha_1} + \ldots + x_{\alpha_n}$.. So $i_\sigma(x)$, for $x \in E_\sigma$ in this light is just the formal sum $x$ (just one term) with $0$ everywhere except $\sigma$.
The function $\pi_\alpha$ goes back to the individual $E_\alpha$ again. For any finite formal sum we can still distinguish the individual vectors (because we never really add them, except within an original vector space), so $\pi_\alpha(x_{\alpha_1} + \ldots + x_{\alpha_n}) = x_{\alpha_i}$ if $\alpha = \alpha_i$ and $0$ if $\alpha \notin \{\alpha_1, \ldots, \alpha_n\}$ (because we work under the idea we have $0$ for every unmentioned index). Formally,when $x$ is a function from $A$ to $\sum_\alpha E_\alpha$, we just define $\pi_\alpha(x) = x(\alpha)$: all elements of $\oplus_\alpha E_\alpha$ are functions from $A$, so we just take the value at $\alpha$, which lies in $E_\alpha$ by condition (i).
So now $\pi_\rho(i_\sigma)$ has two cases: $\sigma \neq \rho$: $i_\sigma(x)$ is the function that sends $\sigma$ to $x$ and all other indices, to $0$. So $i_\sigma(x)(\rho) = 0 = \pi_\rho(i_\sigma(x))$. For $\sigma=\rho$ we just get $x$ again: $\pi_\sigma(i_\sigma(x)) = i_\sigma(x)(\sigma) = x$ by definition. We first make a trivial sum with only $x \in E_\sigma$ and then ask: what is the part from $E_\sigma$? Well, $x$ again, clearly. 
Your $x_\rho$ is just $x(\rho)$, as well. In this setup members of $\oplus_{\alpha \in A} E_\alpha$ are special functions on $A$ (You pick $x(\alpha)$ from $\cup_\alpha E_\alpha$. but you have to choose $0 \in E_\alpha$ almost all of the time, we only have finitely many "real choices" to make, because we only want to do finite formal sums (we could define infinite ones as well). In 2.26 it looks like we do infinite sums, but for every $x$ in $\oplus_\alpha E_\alpha$ the sum is really a finite one.
