# Math Help - Show f is a bijection

1. ## Show f is a bijection

For any p ∈B
f: (A x {p})→ A
Show f is a bijection (one-to-one and onto)

2. ## Re: Show f is a bijection

Originally Posted by Kiefer
For any p ∈B
f: (A x {p})→ A
Show f is a bijection (one-to-one and onto)
You have not defined the function.
You have told us the domain and range.
BUT what is $f(A\times\{p\})=~?$

3. ## Re: Show f is a bijection

This is part of a topology proof: For any p2 ∈ X2
π1|(x1x{p2}): (X1 x {p2})→ X1 is a homeomorphism

4. ## Re: Show f is a bijection

Originally Posted by Kiefer
This is part of a topology proof: For any p2 ∈ X2
π1|(x1x{p2}): (X1 x {p2})→ X1 is a homeomorphism
You cannot just throw us down into the middle of a proof.
If you mean a mapping $f:A\times\{p\}\to A$ defined by $(a,p)\mapsto a$, which is the natural projection of $A\times\{p\}$ onto $A$ then there is nothing to prove. It is a bijection.

5. ## Re: Show f is a bijection

This is part of a topology proof: For any p2 ∈ X2
π1|(x1x{p2}): (X1 x {p2})→ X1 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: (X1 x {p2})→ X1 is a bijection
Proof: Let g=π1|(x1x{p2}) (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∈ X1, there exists p∈ (X1 x {p2}) such that g(p)=x (ie: g is surjective or onto)
Proof: consider any x∈ X1
[Again, this seems like a simple proof, but I could use some help here]
Therefore, g: (X1 x {p2})→ X1 is injective from Claim 1.1
And g: (X1 x {p2})→ X1 is surjective from Claim 1.2
Therefore, g: (X1 x {p2})→ X1 is a bijection

6. ## Re: Show f is a bijection

no, you have to PROVE that if g(x1,p) = g(x2,p) that (x1,p) = (x2,p).

however, note that g(x,p) = x, for all x in X.

so if g(x1,p) = g(x2,p), it is immediate that x1 = x2.

since x1 = x2, and p = p, (x1,p) = (x2,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.