I want to show that given abelian groups $A_i$ for $i\in I$ and $B$, and group homomorphisms $f_i:A_i\rightarrow B$. Let $\iota_i$ denote the inclusion map of $A_i$ into $\bigoplus_{i \in I} A$. Then there exists a unique map $f:\bigoplus_{i \in I} A_i \rightarrow B$ satisfying $f\circ\iota_i=f_i$.
Here is my attempt at proving such a map exists
Claim: The map $f:\bigoplus_{i \in I} A \rightarrow B$ given by $\bigoplus_{i \in I} A \ni \xi=(a_i)_{i\in I}\mapsto \sum_{i\in I}f_i(a_i)$ satisfies $f\circ\iota_i=f_i$.
We note that this sum is well defined since only finitely map $a_i$ are non zero. Let $j\in I$ and $a\in A_j$ be arbitrary. Now $\iota_j(a)=(\xi_i)_{i \in \I}$ where $\xi_i=0_{A_i}$ if $i\neq j$ and $\xi_j=a$. Hence $f\circ\iota_j(a)=\sum_{i\in I,i\neq j}f_i(0_{A_i})+f_j(a)$. Since the $f_i$ are group homomorphisms we conclude $f\circ\iota_j(a)=f_j(a)$.
I'm not sure how to go about proving the uniqueness. If I assume there is a $g$ that satisfies these conditions then it should coincide with $f$ on the individual elements in $A_i$. I think we can use something along the lines of: since $g$ is a group homo we can break it up into how it acts on the different $A_i$ but I am not sure how to do this rigorously.
Best Answer
You're exactly right! You want to show any such $g$ does the same thing as $f$, and this will follow from $g$ being a homomorphism. Formally:
Say $f_i : A_i \to B$, and then define $f : \oplus_i A_i \to B$ as you did.
Now suppose $g : \oplus_i A_i \to B$, and further $g \iota_i = f_i$ for all $i$. We want to show that $f = g$.
Let $x \in \oplus_i A_i$. Then $x = \sum \iota_i a_i$ for some $a_i \in A_i$.
Then $g(x) = g(\sum \iota_i a_i) = \sum g \iota_i a_i = \sum f_i a_i = f(x)$.
Here the first equality is the definition of $x$, the second is because $g$ is a homomorphism,the third is the property we are assuming of $g$, and the last is the definition of $f$.
You had all the right ideas! Writing it down as you said is basically rigorous enough. If you still have questions about why this is rigorous, I'd be happy to answer in the comments.
I hope this helps ^_^