Bijection proof for set products

[The1TL]

Let A1, A2, T be non-empty sets such that A1 is bijective to A2.

Show that A1 × T is bijective to A2 × T



So far I've been able to show that for any b where b is an element of A2, there must be some a within A1 such that f(a) = b. I've been able to do the same for proving injectivity between A1 and A2. I just can't figure out how to apply this to A1 x T and A2 x T.

 

[Petek]

Let $f: A_1 \rightarrow A_2$ be a bijection between $A_1$ and $A_2$. Can you find a bijection between $A_1 \times \ T$ and $A_2 \times \ T$?

 

[The1TL]

exactly, I am pretty sure I need to prove that for some function g: A1 x T --> A2 x T , any (a2, t2) that is an element of A2 x T must have some element (a1,t1) in A1 x T such that g(a1,t1) = (a2,t2). I am just not sure how to show this. I have the same problem with showing the injective aspect of the bijection proof.

 

[Petek]

Here's another hint: If T is any non-empty set, you can always find a bijection between T and itself.

 

[Deveno]

given the bijection f, is k:A1 x T → A2 x T

given by k(a1,t) = (f(a1),t) a bijection?

 

[The1TL]

but is it necessarily true that T must be a bijection to itself?

 

[Deveno]

isn't the function t→t, for all t in T a bijection? let's give it a name, we'll call it g.

so g(t) = t, for all t in T.

is it unclear to you whether or not this is a bijection?

 

[The1TL]

I see how that specific function is a bijection. But I don't see how all functions from T to T would have to be bijections.

 

[Deveno]

we don't need to find "all" bijections between A1 xT and A2 x T. just one will do.