Need some help proving Claim2.1
g1:X1xX2→ X1 and g2:X1xX2→ X2
Claim 2: if (g1 o f): E → X1 and (g2 o f): E → X2 are both continuous, then f: E → (X1 x X2) is continuous.
Proof: Let (g1 o f): E → X1 and (g2 o f): E → X2 both be continuous
We know {(G1xG2)| G1∈Τ1 and G2∈Τ2 } is a base for ΤX1 x X2
Claim 2.1: If A∈B={(G1xG2)| G1∈Τ1 and G2∈Τ2 }, then f-1[A] is open in (E, ΤE)
Proof: Let A∈B
Then G1∈Τ1 (ie: G1 is open)
Then (g1of)-1[G1] ∈ΤE (ie: (g1of)-1[G1] is open) because g1of is continuous
[I'm stuck on how to reach the conclusion from here]
f-1[A] ∈ΤE (ie: f-1[A] is open in (E, ΤE))
Therefore, f: E → (X1 x X2) is continuous
Therefore, f: E → (X1 x X2) is continuous iff (g1 o f): E → X1 and (g2 o f): E → X2 are both continuous from Claims 1 and 2