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The empty-set cases for Cartesian projections matter here. For $p_i$, write $X_j$ for the other factor. If $X_j$ is nonempty, any $x_i\in X_i$ can be paired with an element of $X_j$, so $p_i$ is surjective. If $X_j$ is empty, the Cartesian product is empty and its image is empty. ConsequentlyThe possibility $X_i=\varnothing$ is included because the empty function onto the empty set is surjective.

If $X_i$ is empty, $p_i$ has empty domain and is injective. If $X_j$ has at most one element, an ordered pair is determined by its $i$th coordinate, so $p_i$ is also injective. Conversely, when $X_i$ is nonempty and $X_j$ contains two different elements, pairing those two elements with one fixed $x_i$ gives distinct ordered pairs with the same projection map. Therefore