What is the essential feature that makes the back-and-forth method produce an isomorphism (rather than merely an embedding from A into B)?
AThe method starts from a full isomorphism and restricts to finite pieces, guaranteeing surjectivity throughout
BExtending the partial map in BOTH directions — forward (Spoiler plays in A, Duplicator responds in B) and backward (Spoiler plays in B, Duplicator responds in A) — ensures every element of both structures is eventually mapped, giving surjectivity
CThe method uses the axiom of choice to select a total function simultaneously, ensuring bijectivity
DBecause we are working with countable structures, injectivity automatically implies surjectivity by cardinality
The 'back' direction is what separates back-and-forth from a simple forward construction. A forward-only construction builds an embedding from A to B — every element of A gets a match in B, but elements of B may be left unmatched, so the map need not be surjective. By also going backward (Spoiler names an element of B, Duplicator finds a match in A), every element of B eventually gets matched. The union of all partial maps over infinitely many rounds is then a total bijection — an isomorphism.
Question 2 Multiple Choice
Cantor used the back-and-forth method to prove any two countable dense linear orders without endpoints are isomorphic. What property of dense linear orders ensures the construction never gets stuck?
ADense linear orders are well-ordered, so there is always a minimal element to map next
BBetween any two existing mapped points in a dense order, there is always another element, so a suitable matching element can always be found for any new point
CDense orders have no endpoints, so the mapping can always be extended at the extremes
DCountable orders are isomorphic to ℚ, and ℚ is known to embed in any linear order
Density is the key: for any two elements a < b in a dense linear order, there exists c with a < c < b. When extending the partial map to a new element x (which must fit between two already-mapped elements), density guarantees there is always a suitable match in the other structure — something that fits in the required interval. Without density, the construction could fail if you need to insert a new element between two consecutive elements. Density + no endpoints together ensure every finite partial isomorphism can be extended.
Question 3 True / False
If Duplicator can typically respond in the forward direction (Spoiler plays in A, Duplicator responds in B) for arbitrarily many rounds, the union of the partial maps gives a total isomorphism from A to B.
TTrue
FFalse
Answer: False
Forward-only extendability gives only a total elementary embedding from A to B — every element of A gets matched, but elements of B that are never named by Spoiler may be left unmapped. The resulting map need not be surjective. For a total isomorphism, you need the backward direction too: Spoiler must also be able to play in B, and Duplicator must respond in A. This is the essential 'back' step that drags every element of B into the map, ensuring surjectivity.
Question 4 True / False
A back-and-forth system is a non-empty collection of partial isomorphisms closed under one-step extension in both directions. If the structures are countable, this is sufficient to produce a total isomorphism.
TTrue
FFalse
Answer: True
This is the standard back-and-forth system theorem. Given a non-empty back-and-forth system between countable structures, enumerate all elements of both structures. At each step, use the closure property to extend the current partial map to cover the next unenumerated element (alternating between A and B). The result is a total bijection that is a union of partial isomorphisms — hence itself an isomorphism. The non-emptiness and closure properties convert the local extendability condition into a global construction.
Question 5 Short Answer
Explain why the 'back' direction — extending the partial map from B to A rather than always from A to B — is essential for producing an isomorphism rather than just an embedding.
Think about your answer, then reveal below.
Model answer: An isomorphism must be a bijection: injective (no two elements of A map to the same element of B) and surjective (every element of B is in the image). A forward-only construction handles injectivity and produces an embedding from A into B, but elements of B that are never explicitly matched may be left out of the image. The 'back' step forces every element of B to be named by Spoiler and matched in A, guaranteeing surjectivity. Without the backward extensions, you can build a countable elementary embedding A → B while B has 'extra' elements A doesn't know about — for instance, B could be a proper elementary extension of A. The back direction eliminates this possibility.
This is why the method is called 'back-and-forth' rather than simply 'forth': the backward moves are not optional — they are what turn an embedding into an isomorphism. For proving elementary equivalence (not isomorphism), only finitely many rounds are needed; for building the actual isomorphism, you must iterate infinitely and cover all elements in both directions.