# Multivariable function( 2 dim to 2 dim) set theoretic analysis

Discussion in 'Analysis and Topology' started by deter_gustav01, Oct 20, 2023.

1. ### deter_gustav01

Joined:
Oct 20, 2023
Messages:
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Hi there,

I have this question below and it is about how to prove that the function is ONTO,
With sets, which are arbitrary sets U, V, W, Y.

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Let f:U→V and g:W→Y be onto functions.

Prove that ϕ:U×W→V×Y, defined by ϕ(u,w)=(f(u),g(w)), where
∀u∈U,∀w∈W,
is an onto function.
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My approach was to say let an element, coordinate pair (v,y) ∈ VxY,
And since we are given from the beginning that f and g are ONTO,
that allows one to write that f(u) = v and g(w) = y.
ϕ(u,w)=(v,y).

From this point, I am not certain how to create some
more algebra, to somehow illustrate that ϕ:U×W→V×Y is an ONTO function.

Hope someone can provide some insight into this.

deter_gustav01, Oct 20, 2023
2. ### HallsofIvy

Joined:
Nov 6, 2021
Messages:
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78
To prove that a function, f:X-> Y, s "ONTO" set Y we must show that for any y in Y, there exist x in X such that f(x)= y.

In this case if y is in Vx Y then y= (v, y) with v in V and y in Y. Since f: U-> V is "onto" there exist u in U such that f(u)= v. Since g:W-> Y is "onto" there exist w in W such that g(w)= y. Then (u,w) is in UxW such that fxg(u, w)= (f(u), g(w))= (v, y) so f x g is "onto".

HallsofIvy, Nov 20, 2023
3. ### ROBERTMILLS

Joined:
Feb 27, 2024
Messages:
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To demonstrate that a function f:X→Y is "onto" or surjective onto set Y, we need to prove that for every y in set Y, there exists an x in set X such that f(x)=y.

In this scenario, if y is an element in Y, it can be represented as (v,y) with v in V and y in Y. Given that f:U→V is "onto," there exists an u in U such that f(u)=v. Similarly, as g:W→Y is "onto," there exists a w in W such that g(w)=y. Thus, (u,w) belongs to U×W such that f×g(u,w)=(f(u),g(w))=(v,y), demonstrating that f×g is "onto".

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ROBERTMILLS, Mar 12, 2024