Question about universal property of direct sum of abelian groups

Question

Let $(A_\alpha)_{\alpha \in \Omega}$ be a family of abelian groups, $K$ an abelian group, and $(f_\alpha:A_\alpha\rightarrow K)_{\alpha\in \Omega}$ a family of group morphism.

We define, for each $\alpha \in \Omega$, $i_{\alpha} : A_\alpha \rightarrow \bigoplus_{\gamma\in \Omega}A_\gamma$ by $i_\alpha(a) = (b_\gamma)_{\gamma \in \Omega}$, $b_\alpha = a$, $b_\gamma = 0$ otherwise. Then, there exists a unique morphism $f : \bigoplus_{\alpha\in \Omega}A_\alpha \rightarrow K$ such that $\forall \alpha \in \Omega$, $f\circ i_\alpha=f_\alpha$.

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I understand how to prove this statement. However, what I don't understand is why/how this property determines uniquely (up to an isomorphism) $\bigoplus_{\alpha\in \Omega}A_\alpha$.

What do I have to suppose in order to prove that ? I mean, to prove unicity of something, you have to say there exists something with the same property and show they're the same (isomorphic in that case), but I don't seem to understand how the property here is general. It seems to be linked so much to $\bigoplus_{\alpha\in \Omega}A_\alpha$ itself that I don't know what another group $G$ would have to satisfy in order to prove unicity (isomorphism).

Answer 1

Assume that $G$ satisfies the following (universal) property, call it $(P_G)$ :

For each $\alpha \in \Omega$, there is a group morphism $j_{\alpha} : A_\alpha \rightarrow G$ such that for any abelian group $L$, and any family of group morphisms $(g_\alpha:A_\alpha\rightarrow L)_{\alpha\in \Omega}$, there exists a unique morphism $g : G \rightarrow L$ such that $\forall \alpha \in \Omega$, $g \circ j_\alpha=g_\alpha$.

The goal is to show $$G \cong \bigoplus_{\gamma\in \Omega}A_\gamma =: A$$

You can apply $(P_G)$ to $L=\bigoplus_{\gamma\in \Omega}A_\gamma$ and $g_{\alpha}=i_{\alpha}$. You'll get some group morphism $g : G \rightarrow A$.

Apply the universal property $(P_A)$ of $A$ for $K=G$, and similarly you'll get some group morphism $f : A \to G$. I let you think what you can say about $f \circ g$ and $g \circ f$.

Answer 2

Suppose $G$ is another abelian group, together with morphisms $g_{\alpha}:A_{\alpha}\to G,$ satisfying the universal property. Then we have the two commutative diagrams, for all $\alpha\in \Omega$:

\begin{array}{&&} A_{\alpha } & \stackrel{i_{\alpha }}{\to} & \bigoplus_{\gamma\in \Omega}A_\gamma \\ & g_{\alpha } \searrow & \downarrow \phi \\ & & G \end{array}

and

\begin{array}{&&} A_{\alpha } & \stackrel{g_{\alpha }}{\to} & G \\ & i_{\alpha } \searrow & \downarrow \phi' \\ & & \bigoplus_{\gamma\in \Omega}A_\gamma \end{array}

where $\phi $ and $\phi'$ are the $unique $ morphisms that satisfy

$\tag1 \phi\circ i_{\alpha}=g_{\alpha }$

and

$\tag2 \phi'\circ g_{\alpha }=i_{\alpha}$

Susbtituting $2)$ into $1)$, we get

$\tag3 (\phi\circ \phi')\circ g_{\alpha}=g_{\alpha}$

Now, $3),$ corresponds to the diagram

\begin{array}{&&} A_{\alpha } & \stackrel{g_{\alpha }}{\to} & G \\ & g_{\alpha } \searrow & \downarrow \phi\circ \phi' \\ & & G \end{array}

But we may invoke the UMP one more time to assert that $\phi\circ \phi'=id_G$.

Similarly, if we substitute $1)$ into $2)$, we get, using the same reasoning, $\phi'\circ \phi=id_{\bigoplus_{\gamma\in \Omega}A_\gamma}$, which says $\phi $ is an isomorphism.