Universal Property of Products in $A$-modules

Question

I am studying some notes in which the second Universal Property of Products in $A$-modules is defined as following:

Given an $A$-modules family $\{M_i\}_{i\in I}$, for all test $A$-module $T$ we have a canonical isomorphism:

$$ \Pi_{i\in I}\operatorname{Hom}_A(M_i,T)=\operatorname{Hom}_A(\oplus_{i\in I}M_i,T)\\ (\psi_i)_{i\in I}\mapsto ((m_i)_{i\in I}\to \sum _{i\in I}\psi_i(m_i)).$$

My question, maybe trivial (but I am stuck on that...), is the following: suppose that $I$ is infinite and $\psi_i$ non-zero are infinite. How to choose the subfamily $(m_i)\in \oplus_{i\in I}M_i$ to construct the image of the isomorphism once there are infinites subfamilies $(m_i)\in \oplus_{i\in I}M_i$ (i.e., with almost $m_i=0$)?

Many thanks in advance for some help!

Answer 1

Na'omi kindly invited me to post the following comment as an answer.

Here is a slightly more explicit statement of the universal property in question:

Given an $A$-modules family $\{M_i\}_{i\in I}$, for all test $A$-module $T$ we have a canonical isomorphism: $$ \Phi:\prod_{i\in I}\text{Hom}_A(M_i,T)\to\text{Hom}_A\left(\bigoplus_{i\in I}M_i,T\right) $$ defined by $$ (\Phi((\psi_i)_{i\in I}))((m_i)_{i\in I})=\sum _{i\in I}\psi_i(m_i). $$

Answer 2

The subtlety is that in $(m_i)_{i \in I} \in \bigoplus_{i \in I} M_i$ we have $m_i = 0$ for all but finitely many $i \in I$ by construction of the direct sum. That's why the sum $\sum_{i \in I} \psi_i(m_i)$ is well-defined: almost every term is $0$.