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