For any p ∈B
f: (A x {p})→ A
Show f is a bijection (one-to-one and onto)
This is part of a topology proof: For any p_{2} ∈ X_{2}
π_{1}|(x_{1}x{p_{2}}): (X_{1} x {p_{2}})→ X_{1} is a homeomorphism
Sorry, I phrased that poorly. Above is the whole topology proof as given to me in class. In my original post, I was just trying to simplify notation. Part of being a homeomorphism is being a bijection, and that is the part I need help proving. Here is an outline for the proof:
Proof: Consider any p2 ∈ X2
Claim 1: g: (X_{1} x {p_{2}})→ X_{1} is a bijection
Proof: Let g=π_{1}|(x_{1}x{p_{2}}) (just to simplify notation)
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)
[Can I simply say g(g(p))=g(g(q)) and therefore 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
no, you have to PROVE that if g(x_{1},p) = g(x_{2},p) that (x_{1},p) = (x_{2},p).
however, note that g(x,p) = x, for all x in X.
so if g(x_{1},p) = g(x_{2},p), it is immediate that x_{1} = x_{2}.
since x_{1} = x_{2}, and p = p, (x_{1},p) = (x_{2},p) by the definition of equality in the cartesian product.
you also have to prove that for ANY x in X, there is some y in X x {p} with g(y) = x.
i suggest y = (x,p).
**********
the fact that g:X x {p} → X given by g(x,p) = x is a bijection is, admittedly, rather trivial. p is just "along for the ride", all we do for any x in X is just add (_,p) to it.
the proof that g is a HOMEOMORPHISM is more involved. here, the explict TOPOLOGIES on Xx{p} and X need to be SPECIFIED.
presumably, you are using the subspace (relative) topology of Xx{p} relative to the PRODUCT topology on XxY (where p is a point of Y), where some explicit topologies are given on X and Y.
although you do not say so, it may be the case that X = Y = R, the real numbers with the standard topology (defined by the base of open intervals under the euclidean metric).
in any case, it IS clear that g is a continuous bijection. what is less trivial is that g^{-1} is also continuous.