Let f: A → B be a function, and let E and F be subsets of A. Consider the…

For a function f: A → B and subsets E, F of A, the image of a set S under f is f(S) = {f(x) : x ∈ S}. Two general facts govern how images interact with union and intersection: images always distribute over a union — f(E ∪ F) = f(E) ∪ f(F) for every f, E, F — but images distribute over an intersection only when f is injective; in general only the inclusion f(E ∩ F) ⊆ f(E) ∩ f(F) is guaranteed, and it can be strict.

Checking S₁ — image of a union:

1. Let y ∈ f(E ∪ F). Then y = f(x) for some x ∈ E ∪ F.

2. So x ∈ E or x ∈ F, which makes y ∈ f(E) or y ∈ f(F), i.e. y ∈ f(E) ∪ f(F). This gives f(E ∪ F) ⊆ f(E) ∪ f(F).

3. Conversely, if y ∈ f(E) ∪ f(F), then y ∈ f(E) or y ∈ f(F), so y = f(x) for some x ∈ E or x ∈ F, i.e. x ∈ E ∪ F, so y ∈ f(E ∪ F). This gives f(E) ∪ f(F) ⊆ f(E ∪ F).

4. Both inclusions use only the definition of image and union, with no restriction on f — so S₁ holds for every function and every E, F.

Checking S₂ — image of an intersection:

1. The inclusion f(E ∩ F) ⊆ f(E) ∩ f(F) always holds, by the same forward argument as above.

2. For the reverse inclusion to hold, no two distinct elements of A may share an image on the relevant sets. If f is not injective, pick a ≠ b in A with f(a) = f(b), and set E = {a}, F = {b}.

3. Then E ∩ F = ∅, so f(E ∩ F) = ∅.

4. But f(E) ∩ f(F) = {f(a)} ∩ {f(b)} = {f(a)}, since f(a) = f(b) — a non-empty set.

5. So f(E ∩ F) ≠ f(E) ∩ f(F) for this choice — S₂ fails whenever f is not injective, so it is not true in general.

Cross-check with concrete sets: let A = {1, 2}, B = {0}, f(1) = f(2) = 0, E = {1}, F = {2}. Then E ∩ F = ∅ gives f(E ∩ F) = ∅, while f(E) ∩ f(F) = {0} ∩ {0} = {0} — confirming S₂ breaks down exactly as predicted, while the S₁ proof used no assumption on f and so is unaffected by this or any other example.

Only S₁ is universally true; S₂ additionally requires f to be injective. So the correct choice is the option asserting only S₁ holds.