Q:

Direct product of injective modules is injective: stucked in proof

This is very simple exercise in homological algebra texts, but I was trying to prove it myself, which I couldn't do by one stuck.  
Claim: If $I_j$ is injective then $\prod_{j\in J} I_j$ is injective.  
(1)  Let $\alpha: A\rightarrow \prod_j I_j$ a homomorpism, $\mu:A\rightarrow B$ a monomorphism, $p_j:\prod_j I_j\rightarrow I_j$ projections..
(2) By injectivity of $I_j$, there exists unique $\beta_j:B\rightarrow I_j$ s.t.  $\beta_j\mu=p_j\alpha$.
(3) By universal property of product $\prod_j I_j$, there exists unique $\beta:B\rightarrow \prod_j I_j$ s.t.  $\beta_j=p_j\beta$. Thus $$\beta_j\mu=p_j\alpha \hskip5mm\mbox{  i.e. }\hskip5mm p_j\beta\mu=p_j\alpha.  $$ 
I coundn't proceed further to conclude that $\beta\mu=\alpha$. How can I do this?  

A:

The universal property of products tells you that if $p_jf=p_jg$ for all $j\in J$, then $f=g$. Since $p_j(\beta\mu)=p_j\alpha$, you're done.  Your condition $(3)$ is (emphasis on unique added)

By the universal property of product $\prod_j I_j$, there exists a unique $\beta:B\rightarrow \prod_j I_j$ such that $\beta_j=p_j\beta$. Thus $$\beta_j\mu=p_j\alpha \qquad\text{i.e.}\qquad p_j\beta\mu=p_j\alpha.  $$ 

