f: E → (X1 x X2) is cont iff (g1 o f): E → X1 and (g2 o f): E → X2 are both cont

I have an outline for the proof, but would appreciate someone checking if the logic in Claim 1.1 is sound and I could use some help with Claim 2.1

For any topological space (E, Τ_{E}) and any function f: E → (X_{1} x X_{2}),

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.

Proof:

Claim 1:

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

Proof: Suppose f: E → (X_{1} x X_{2}) is continuous.

Claim 1.1: (g_{1} o f): E → X_{1} is continuous

Suppose G∈Τ_{1} (ie: G is open in X_{1})

Then G∈Τ_{X1 x X2} [Is this notation ok for the topology associated with X_{1} x X_{2}? Is this line true? Do we know Τ_{1} ⊆ Τ_{X1 x X2}?]

Then f^{-1}[G] ∈Τ_{E} because f: E → (X_{1} x X_{2}) is continuous

Then (g_{1} o f)^{-1}[G] ∈Τ_{E} (ie: (g_{1} o f) is open) [Do we know this from the above step?]

Hence, (g_{1} o f) is continuous

Similarly, we can show (g_{2} o f) is continuous

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