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The universal property of Cartesian products constructs the desired map pointwise:Its coordinate projection maps are $f_1$ and $f_2$. Any other map with those projection maps must have the same two coordinates at every $y$, and hence must equal $g$. This also covers $Y=\varnothing$.

For joint injectivity of a pair of functions, equality $g(y)=g(y^{\prime })$ means equality of both coordinate images. If either coordinate function is injective, it follows that $y=y^{\prime }$, so $g$ is injective. The converse fails: take $Y=\{a,b,c\}$ and map these points to $(0,0),(0,1),(1,0)$ in $\{0,1{\}}^2$. The paired map is injective, but the first coordinate identifies $a,b$ and the second identifies $a,c$.

There is an empty-factor defect in the printed surjectivity implication. If both $X_1$ and $X_2$ are nonempty and $g$ is surjective, every element of either factor can be completed to a pair and then lifted through $g$. Hence both $f_i$ are surjective. If both factors are empty, existence of the $f_i$ forces $Y$ to be empty and both coordinate maps are again surjective. But takeThen $g:\varnothing \to X_1\times X_2=\varnothing$ is surjective, while $f_1:\varnothing \to \{0\}$ is not. Without a nonempty-factor assumption, the printed implication is false. This is exactly the exception found for the projection maps in part (ii).

Even when both factors are nonempty, the reverse implication fails. For $Y=X_1=X_2=\{0,1\}$, let $f_1(y)=f_2(y)=y$. Both coordinate maps are surjective, but $g(y)=(y,y)$ misses $(0,1)$ and $(1,0)$. Thus coordinate surjectivity does not guarantee that every pair is attained.