Suppose that $f_\alpha : X\longrightarrow Y_\alpha$ for any $\alpha\in I$, where $X$,$Y_\alpha$ be nonempty sets for all $\alpha\in I$, and Let $f : X\longrightarrow \prod\{Y_\alpha: \alpha\in I \}$ be the map defined by $f(x)(\alpha)= f_\alpha(x)$ for any $x\in X$.
The map $f$ is called the diagonal product of the family $\mathfrak{F}=\{f_\alpha:\alpha\in I\}$.
Is it true that $f$ is injective?
Clearly $f(x)=f(y)$ if and only if $f_\alpha(x)=f_\alpha(y)$ for all $\alpha\in I$. It is certainly possible that there are distinct $x,y\in X$ such that $f_\alpha(x)=f_\alpha(y)$ for all $\alpha\in I$, so it is possible that $f$ is not injective.
This is sometimes expressed by saying that $f$ is injective if and only if the family $\mathfrak{F}$ separates point of $X$, meaning that whenever $x,y\in X$ and $x\ne y$, there is an $\alpha\in I$ such that $f_\alpha(x)\ne f_\alpha(y)$.