Need some help proving Claim2.1

g_{1}:X_{1}xX_{2}→ X_{1}and g_{2}:X_{1}xX_{2}→ X_{2}

Claim 2: if (g_{1}o f): E → X_{1}and (g_{2}o f): E → X_{2}are both continuous, then f: E → (X_{1}x X_{2}) is continuous.

Proof: Let (g_{1}o f): E → X_{1}and (g_{2}o f): E → X_{2}both be continuous

We know {(G_{1}xG_{2})| G_{1}∈Τ_{1}and G_{2}∈Τ_{2}} is a base for Τ_{X1 x X2}

Claim 2.1: If A∈B={(G_{1}xG_{2})| G_{1}∈Τ_{1}and G_{2}∈Τ_{2}}, then f^{-1}[A] is open in (E, Τ_{E})

Proof: Let A∈B

Then G_{1}∈Τ_{1 }(ie: G_{1}is open)

Then (g_{1}of)^{-1}[G_{1}] ∈Τ_{E}(ie: (g_{1}of)^{-1}[G_{1}] is open) because g_{1}of is continuous

[I'm stuck on how to reach the conclusion from here]

f^{-1}[A] ∈Τ_{E}(ie: f^{-1}[A] is open in (E, Τ_{E}))

Therefore, f: E → (X_{1}x X_{2}) is continuous

Therefore, f: E → (X_{1}x X_{2}) is continuous iff (g_{1}o f): E → X_{1}and (g_{2}o f): E → X_{2}are both continuous from Claims 1 and 2