I'm having trouble with the following proposition:
I'm assuming that under the function $f$, an element $(e_i)_{i \in I}$ of $\bigoplus_{i \in I} E_i$ gets mapped to $\sum_{i \in I} f_i(e_i).$ In this case, the author claims that the conditions are obviously necessary for they are satisfied by the $g_i = pr_i \circ f^{-1}$. However, I don't see how the function $g_i$ satisfies the first property, for instance. Indeed, let $e_i \in E_i$. Then $f_i$ maps $e_i$ to some element of $F$ and supposedly (I don't understand this part exactly) $f^{-1}$ takes this very element of $F$ and maps it back to the element $(0,...,0,e_i,0,...)$ of $\bigoplus_{i \in I} E_i$, which the projection map then sends back to $e_i$.
If anyone could clarify the part I mentioned I don't understand, I'd appreciate it a lot. Thanks.
In case anyone was wondering, this is from Bourbaki's Algebra I, page 203.
Well, clearly $(0_1, \ldots, 0_{i - 1}, e_i, 0_{i + 1}, \ldots)$ is mapped by $f$ to $$ f_1(0_1) + \ldots + f_{i - 1}(0_{i - 1}) + f_i(e_i) + f_{i + 1}(0_{i + 1}) + \ldots = f_i(e_i) $$ So if $f$ is an isomorphism, thus bijective, this must be the only element of $\bigoplus_{i \in I} E_i$ getting mapped to $f_i(e_i)$ i.e. $f^{-1}(f_i(e_i)) = (0_1, \ldots, 0_{i - 1}, e_i, 0_{i + 1}, \ldots)$.