Direct product of injective modules is injective: stucked in proof

Question

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?

Answer 1

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 $$\beta_j\mu=p_j\alpha \qquad\text{i.e.}\qquad p_j\beta\mu=p_j\alpha. $$