Prove universal property of direct sum Let $\{X_\alpha\}_{\alpha\in A}$ be a collection of groups and let $X$ be a group such that for all $\alpha \in A$ there exists an injection $i_\alpha \colon X_\alpha\hookrightarrow X$. Furthermore suppose that $X$ has the following property: For all groups $Y$ with a collection of homomorphisms $\{f_\alpha\}_{\alpha\in A}$ where $f_\alpha \colon X_\alpha\to Y$ for $\alpha\in A$ there exists a unique homomorphism $f \colon X\to Y$ with the property $f\circ i_\alpha=f_\alpha$ for all $\alpha\in A$. Prove that $X$ is isomorphic to $\bigoplus_{\alpha\in A}X_\alpha$.
Here is what I have done:
Let $Y=\bigoplus_{\alpha\in A}X_\alpha$ and for $\alpha\in A$ we define $f_\alpha:X_\alpha\to Y$ by $f_\alpha(x_\alpha)=(f^\beta_\alpha(x_\alpha))_{\beta\in A}$ where $f^\beta_\alpha(x_\alpha)=e_{\beta}$ (where $e_\beta$ is the neutral element of $X_\beta$) if $\beta\neq\alpha$ and $f^\beta_\alpha(x_\alpha)=x_\alpha$ if $\beta=\alpha$ for $x_\alpha\in X_\alpha$. We see that $\{f_\alpha\}_{\alpha\in A}$ is a family of homomorphisms, so by hypothesis there exists a unique homomorphism $f:X\to Y$ with $f\circ i_\alpha=f_\alpha$ for all $\alpha\in A$. So we try to show that $f$ is an isomorphism.
Surjectivity:
Let $(x_\alpha)_{\alpha\in A}\in \bigoplus_{\alpha\in A}X_\alpha$ with $x_\alpha\ne e_\alpha$ if and only if $\alpha\in\{\alpha_1,\ldots,\alpha_r\}$ for an $r\in\mathbb{N}_0$. Now consider $x=i_{\alpha_1}(x_{\alpha_1})\cdot\ldots\cdot i_{\alpha_r}(x_{\alpha_r})$; we see that $f(x)=f\left(i_{\alpha_1}(x_{\alpha_1})\cdot\ldots\cdot i_{\alpha_r}(x_{\alpha_r})\right)=f\left(i_{\alpha_1}(x_{\alpha_1})\right)\cdot\ldots\cdot f\left(i_{\alpha_r}(x_{\alpha_r})\right)=f_{\alpha_1}(x_{\alpha_1})\cdot\ldots\cdot f_{\alpha_r}(x_{\alpha_r})=(x_\alpha)_{\alpha\in A}$. Thus $f$ is surjective.
Injectivity:
Equivalently we try to show that $\ker f=\{e_X\}$. We see that if $x\in\langle i_\alpha(X_\alpha)\ \mid\ \alpha\in A\rangle$ then $f(x)=e_X\implies x=e_X$. However, I fail to see how to argue if $x\notin\langle i_\alpha(X_\alpha)\ \mid\ \alpha\in A\rangle$. I think it has something to do with the uniqueness of the homomorphism $f \colon X\to Y$, as so far I haven't used this property.
Am I on the right track? Is what I have done so far correct? How to prove the injectivity? Thanks.
 A: I'm not quite sure why you're doing it this way, and I'm sorry but I don't really feel like checking the correctness of your approach, by checking that the homomorphism you obtained is injective and surjective. To me it looks like a more direct way to show they're isomorphic, but with the notation etc. it's a bit too much given my capabilities, so maybe someone else will check your approach there.
Why not show that $\bigoplus_{\alpha\in A}X_\alpha$ has the universal property, and then show that it's isomorphic to $X$ ?
Then, if you know $\bigoplus_{\alpha\in A}X_\alpha$ has the universal property, it's not that much work to see that it's isomorphic to $X$ (as $X$ also has the universal property). Using 2 objects for 'clarity': 
Let $X_1$ and $X_2$ be two groups, $i_1$ and $i_2$ be the injections to $X$ (presumed to exist, from the original post), and $j_1$ and $j_2$ be the injections to $\bigoplus_{\alpha\in A}X_\alpha$, where $A = \{1,2\}$.
There exists a unique morphism $g$ from $X$ to $\bigoplus_{\alpha\in A}X_\alpha$ satisfying $g \circ i_1 = j_1$ and $g \circ i_2 = j_2$. 
Similarly, there exists a unique morphism $h$ from $\bigoplus_{\alpha\in A}X_\alpha$ to $X$, satisfying $h \circ j_1 = i_1$ and $h \circ j_2 = i_2$.
Using associativity, you can check that $(h \circ g) \circ i_1 = i_1$ and $(h \circ g) \circ i_2 = i_2$, but there's only 1 morphism from $X$ to $X$ that satisfies that (using the universal property of $X$)- identity. So $h \circ g = id_X$. 
Repeat the same argument to show that $g \circ h = id_{\bigoplus_{\alpha\in A}X_\alpha}$.
