I've outlined the proof but could use some help with the details in the simpler sections (eg: proving a bijection)
For any p_{2} ∈ X_{2}
π_{1}|_{(x1x{p2})}: (X_{1} x {p_{2}})→ X_{1} is a homeomorphism (ie: bijective, continuous and open)
Proof: Consider any p_{2} ∈ X_{2}
Let g=π_{1}|_{(x1x{p2})}
Claim 1: g: (X_{1} x {p_{2}})→ X_{1} 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∈ X_{1}, there exists p∈ (X_{1} x {p_{2}}) such that g(p)=x (ie: g is surjective or onto)
Proof: consider any x∈ X_{1}
[Again, this seems like a simple proof, but I could use some help here]
Therefore, g: (X_{1} x {p_{2}})→ X_{1} is injective from Claim 1.1
And g: (X_{1} x {p_{2}})→ X_{1} is surjective from Claim 1.2
Therefore, g: (X_{1} x {p_{2}})→ X_{1} is a bijection
Claim 2: g: (X_{1} x {p_{2}})→ X_{1} is continuous
Proof: We know f_{1}: (X_{1} x X_{2}) → X_{1} is a continuous open surjection
Therefore, g: (X_{1} x {p_{2}})→ X_{1} is continuous [Can we say this follows directly because p_{2} ∈ X_{2} or is there a way to add more detail here?]
Claim 3: If B={(G_{1}x{p_{2}})| G_{1}∈Τ_{1}} is a base for Τ_{X1 x X2}|_{X1 x {p2}} then f_{1}[(G_{1} x {p_{2}})]∈ Τ_{X1 x X2}|_{X1}
Proof: Let B be a base for Τ_{X1 x X2}|_{X1 x {p2}}
Then f_{1}^{-1}[(G_{1} x {p_{2}})]∈ Τ_{X1 x X2}|_{X1 x {p2}}
Then f_{1}[(G_{1} x {p_{2}})]∈ Τ_{X1 x X2}|_{X1}
Ie: g: (X_{1} x {p_{2}})→ X_{1} is open
So g: (X_{1} x {p_{2}})→ X_{1} is a homeomorphism from claims 1,2 and 3