# Thread: Topology Proof: g:(X1 x {p2})→ X1 is a homeomorphism

1. ## Topology Proof: g:(X1 x {p2})→ X1 is a homeomorphism

I've outlined the proof but could use some help with the details in the simpler sections (eg: proving a bijection)

For any p2 ∈ X2
π1|(x1x{p2}): (X1 x {p2})→ X1 is a homeomorphism (ie: bijective, continuous and open)

Proof: Consider any p2 ∈ X2
Let g=π1|(x1x{p2})
Claim 1: g: (X1 x {p2})→ X1 is a bijection
Claim 1.1: If g(p)=g(q), then p=q (ie: g is injective or one-to-one)
Proof: Let g(p)=g(q)
[I know this is simple, but I haven’t done this sort of proof in a while…how do I show p=q? Can I say g(g(p))=g(g(q))? Then p=q?]
Claim 1.2: For every x∈ X1, there exists p∈ (X1 x {p2}) such that g(p)=x (ie: g is surjective or onto)
Proof: consider any x∈ X1
[Again, this seems like a simple proof, but I could use some help here]
Therefore, g: (X1 x {p2})→ X1 is injective from Claim 1.1
And g: (X1 x {p2})→ X1 is surjective from Claim 1.2
Therefore, g: (X1 x {p2})→ X1 is a bijection

Claim 2: g: (X1 x {p2})→ X1 is continuous
Proof: We know f1: (X1 x X2) → X1 is a continuous open surjection
Therefore, g: (X1 x {p2})→ X1 is continuous [Can we say this follows directly because p2 ∈ X2 or is there a way to add more detail here?]

Claim 3: If B={(G1x{p2})| G1∈Τ1} is a base for ΤX1 x X2|X1 x {p2} then f1[(G1 x {p2})]∈ ΤX1 x X2|X1
Proof: Let B be a base for ΤX1 x X2|X1 x {p2}
Then f1-1[(G1 x {p2})]∈ ΤX1 x X2|X1 x {p2}
Then f1[(G1 x {p2})]∈ ΤX1 x X2|X1
Ie: g: (X1 x {p2})→ X1 is open
So g: (X1 x {p2})→ X1 is a homeomorphism from claims 1,2 and 3