Compute f°g & g°f: Domain & Codomain Explained

[Austin Chang]

Homework Statement

Let f: ℝxℝ → ℝ and g: ℝ → ℝxℝ be given by f(x,y) = x and g(x) = (x,0) Compute f°g and g°f (This means for each function find the domain and codomain, and show what each function evaluates at each point in the domain.

Homework Equations

The Attempt at a Solution

g°f ⇒ f(g(x)) or f(x,y) = x → g(x) = (x,0). Therefore x ∈ ℝ and y ∈ ℝ is the domain and x ∈ ℝ and y ∈ 0 is part of the codomain. When placing x ∈ ℝ and y ∈ ℝ in the function the output will be (x,0) iff (x,y)[/B]
That is for g°f. Not going to spend another 3 minutes typing out f°g because I think it would be a waste of mine and your time. I am just really confused about everything on this section as a whole and hopefully your insights will help. Please ... Please point out anything that you don't like about the way I write or my notation or really anything. Thanks

 

[SammyS]

Homework Statement

Let f: ℝxℝ → ℝ and g: ℝ → ℝxℝ be given by f(x,y) = x and g(x) = (x,0) Compute f°g and g°f (This means for each function find the domain and codomain, and show what each function evaluates at each point in the domain.

Homework Equations

The Attempt at a Solution

g°f ⇒ f(g(x)) or f(x,y) = x → g(x) = (x,0). Therefore x ∈ ℝ and y ∈ ℝ is the domain and x ∈ ℝ and y ∈ 0 is part of the codomain. When placing x ∈ ℝ and y ∈ ℝ in the function the output will be (x,0) iff (x,y)[/B]
That is for g°f. Not going to spend another 3 minutes typing out f°g because I think it would be a waste of mine and your time. I am just really confused about everything on this section as a whole and hopefully your insights will help. Please ... Please point out anything that you don't like about the way I write or my notation or really anything. Thanks

g_°f is often expressed as g(f( argument of f )) in this case the argument of f is an element of ℝ², often written as an ordered pair .

f(g(x)) gives f_°g rather than g_° .

g(x) = (x,0) .
Therefore, f(g(x)) = f((x,0)) = x . Your result was correct for f_°g, but the intermediate step was wrong.

 

[PeroK]

I am just really confused about everything on this section as a whole and hopefully your insights will help. Please ... Please point out anything that you don't like about the way I write or my notation or really anything. Thanks

This may or may not help.

Think of a function as a process that takes an input and changes it to an output according to some rule. And, the composition of two functions is like a mini production line.

For this problem you could imagine that you are $f$ and a friend is $g$. If I give you an input $(x, y)$ you transform that into an output according to the rules of $f$ and pass the output to your friend. They apply the rules of $g$ to that. And, as there only these two steps in the production line, that's the final output which gets handed back to me.

So, if I give you $(x,y)$ what do I get back from your friend?

That's g_°f. Which means $f$ first then $g$. You'll have to remember this.

We can also set it up the other way round. For f_°g, I give $x$ to your friend who does $g$ and passes the result to you, who does $f$ and gives the final output to me.

So, if I give your friend $x$ what do I get back from you.

Finally, I never liked the term codomain. I use the terms domain and range. In any case, these define the input and output for a function. The set of all allowed inputs is the domain. And the set of all possible outputs is the codomain /range.

In this case, $f$ takes ordered pairs of real numbers and outputs a single real number, while $g$ does the opposite.

 

[Austin Chang]

Is this true: f is surjective, g is injective, g°f is neither,f°g is bijective?

 

[PeroK]

Is this true: f is surjective, g is injective, g°f is neither,f°g is bijective?

What do you think?

 

[Austin Chang]

Do you want me to explain why i think it is true?

 

[PeroK]

Do you want me to explain why i think it is true?

Yes, that's all true. Although, f o g is more than a bijection: It's the identity function.

 

[Austin Chang]

Do you want me to explain why i think it is true?

Yes, that's all true. Although, f o g is more than a bijection: It's the identity function.

Thank you