isomorphic mapping on direct sum and products I found an old post here: https://www.physicsforums.com/threads/question-about-isomorphic-mapping-on-direct-sums.709423/
While reading the answer to the question posted in the link above, I found that it is conflicting with what I saw here: http://www.math.harvard.edu/~elkies/M55a.10/lemma3.pdf
So the question is: which is true? Is it possible to construct a meaningful isomorphism map in the case of infinite direct sum? In case of direct product?
 A: There are two different claims being made here. One is that: for any $v\in \bigoplus V_i=:V,$ $\sum_i \iota_i\pi_i v=v$. This makes sense, because $\pi_i v=0$ for all but finitely many $i$, and is also true. One is that $\sum_i \iota_i\pi_i=1_{\oplus V_i}$. To the extent that this means the same thing as the first claim, of course it's true. But, and this is Elkies' point, an infinite sum like $\sum_i \iota_i\pi_i$ in a vector space like $\text{End}(\oplus V_i)$ is only defined if all but finitely many of the summands are zero, and all the $\iota_i\pi_i$ are nonzero, so this sum is undefined.
As the physics forum answerer says, it's possible to say that by $\sum \iota_i\pi_i$ I just means the thing that sends $v$ to $\sum\iota_i\pi_i v$, and so the former is defined because the latter is. But this uses more than the vector space structure on $\text{End}V$. Specifically, it uses the fact that $\text{End} V$ acts on $V$, in which case what we mean by the symbol $\sum \iota_i\pi_i$ is not an element of $\text{End} V$ at all, but rather simply a function from $V$ to $V$, constructed out of the natural action of $\text{End} V$ on $V$, which happens to coincide with $1_V$. So the most exact answer is that if you view $\text{End} V$ as an abstract vector space, then the proposed identity is meaningless, but that if view view it as a ring with an associated action on $V$, then the identity holds. So, regardless, the infinite case is clearly distinguished from the finite case, when $\sum \iota_i\pi_i=1_V$ is a perfectly valid identity without any need for $V$ at all.
This is all rather subtle, and probably not very significant in practice, but on the other hand I do think it clarifies the conceptual situation enough to make it worth spelling out.
Regarding direct products, there's no hope: the image of elements of the $V_i$ under the inclusion $V_i\to \prod V_i$ do not even span $\prod V_i$ when there are infinitely many $i$. (If this isn't clear to you, think about the span of the images of copies of $\mathbf{R}$ in the product space $\mathbf{R}^{\mathbf{R}}$.)
