For any p ∈B

f: (A x {p})→ A

Show f is a bijection (one-to-one and onto)

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- Oct 20th 2012, 01:49 PMKieferShow f is a bijection
For any p ∈B

f: (A x {p})→ A

Show f is a bijection (one-to-one and onto) - Oct 20th 2012, 01:56 PMPlatoRe: Show f is a bijection
- Oct 20th 2012, 02:15 PMKieferRe: Show f is a bijection
This is part of a topology proof: For any p

_{2}∈ X_{2 }

π_{1}|_{(x1x{p2})}: (X_{1}x {p_{2}})→ X_{1}is a homeomorphism - Oct 20th 2012, 02:28 PMPlatoRe: Show f is a bijection
You cannot just throw us down into the middle of a proof.

If you mean a mapping $\displaystyle f:A\times\{p\}\to A$ defined by $\displaystyle (a,p)\mapsto a$, which is the natural projection of $\displaystyle A\times\{p\}$ onto $\displaystyle A$ then there is nothing to prove. It is a bijection. - Oct 21st 2012, 05:43 AMKieferRe: Show f is a bijection
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 - Oct 22nd 2012, 09:39 AMDevenoRe: Show f 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.