Abstract - one to one and onto questions

[kathrynag]

Homework Statement

Determine whether the given function is one to one and whether it is onto. If the function is both one to one and onto, find the inverse of the function.
f:$$R^{2}$$$$\rightarrow$$$$R^{2}$$, f(x,y)=(x+y, y) .

Homework Equations

The Attempt at a Solution

I know one to one says f(x)=f(y) implies x=y
Onto means if for every element y in $$R^{2}$$, there exists an element x in $$R^{2}$$ with f(x)=y.
I conceptually understand the idea, but don't know how to use these definitions.

Homework Statement

Let S={1,2,3} and T={4,5}
I need to find how many functions are there from S into T? T into S? And how many of there are one to one and onto.

Homework Equations

The Attempt at a Solution

My problem with this is getting a function from a set.

 

[vela]

Homework Statement

Determine whether the given function is one to one and whether it is onto. If the function is both one to one and onto, find the inverse of the function.
f:$$R^{2}$$$$\rightarrow$$$$R^{2}$$, f(x,y)=(x+y, y) .

Homework Equations

The Attempt at a Solution

I know one to one says f(x)=f(y) implies x=y
Onto means if for every element y in $$R^{2}$$, there exists an element x in $$R^{2}$$ with f(x)=y.
I conceptually understand the idea, but don't know how to use these definitions.

You may find it a bit less confusing if you use letters other than x and y. For the function you've been given, you're trying to show that if f(a)=f(b), then a=b, where a=(x₁,y₁) and b=(x₂,y₂). Similarly, to show it's surjective, you want to show if f(a)=b, you can solve for a. Just write down the two equations each vector equation represents in terms of the components of a and b and solve them.

 

[Hurkyl]

I know one to one says f(x)=f(y) implies x=y
...
I conceptually understand the idea, but don't know how to use these definitions.

Suppose s and t are elements of R² such that f(s) = f(t)
...
...
...
And so s = t.

Therefore, f is one-to-one.

Presumably, you'd use the definition of f somewhere in there, and of R².

 

[vela]

Homework Statement

Let S={1,2,3} and T={4,5}
I need to find how many functions are there from S into T? T into S? And how many of there are one to one and onto.

Homework Equations

The Attempt at a Solution

My problem with this is getting a function from a set.

From a set theoretic point of view, a function is just a list of ordered pairs (x,y) where x is an element of S and y is an element of T. So start by writing down a list of all possible ordered pairs you can have. These are the elements of the cartesian product SxT. A function from S into T will be a subset of that set, so the problem is figuring out how many subsets you can find of SxT.

 

[kathrynag]

You may find it a bit less confusing if you use letters other than x and y. For the function you've been given, you're trying to show that if f(a)=f(b), then a=b, where a=(x₁,y₁) and b=(x₂,y₂). Similarly, to show it's surjective, you want to show if f(a)=b, you can solve for a. Just write down the two equations each vector equation represents in terms of the components of a and b and solve them.

So f(x)=x+y
and f(y)=y
Now to show f(a)=b
f(a)=a+y says f(a)-a=y
f(a)=y
So it is onto?

 

[kathrynag]

From a set theoretic point of view, a function is just a list of ordered pairs (x,y) where x is an element of S and y is an element of T. So start by writing down a list of all possible ordered pairs you can have. These are the elements of the cartesian product SxT. A function from S into T will be a subset of that set, so the problem is figuring out how many subsets you can find of SxT.

Oh ok so something like (1,4) is one function?

 

[vela]

So f(x)=x+y
and f(y)=y
Now to show f(a)=b
f(a)=a+y says f(a)-a=y
f(a)=y
So it is onto?

No, f(x) and f(y) don't make sense because x and y are not elements of R². What are the domain and codomain of f? What do elements of the domain and codomain look like?

 

[vela]

From a set theoretic point of view, a function is just a list of ordered pairs (x,y) where x is an element of S and y is an element of T. So start by writing down a list of all possible ordered pairs you can have. These are the elements of the cartesian product SxT. A function from S into T will be a subset of that set, so the problem is figuring out how many subsets you can find of SxT.

I was too sloppy. It's not all subsets of SxT but only those which satisfy the definition of a function, namely that it's single-valued.

Oh ok so something like (1,4) is one function?

Suppose you have a domain S={1,2,3} and codomain T={a,b,c}. The cartesian product SxT is the set consisting of all ordered pairs: SxT={(x,y) | x is in S and y is in T}. In this case, you'd have

SxT={(1,a), (1,b), (1,c), (2,a), (2,b), (2,c), (3,a), (3,b), (3,c)}

A function f from S to T is a subset of SxT which satisfies the requirement that an element in S maps to one value in T. For example, if you had

f={(1,a), (2,b), (3,c)}

this would be the function where, in more common notation, f(1)=a, f(2)=b, and f(3)=c. On the other hand, the subset

g={(1,a), (1,b)}

is not a function because it would mean that g(1)=a and g(1)=b, which you know isn't true for a function.

 

[kathrynag]

No, f(x) and f(y) don't make sense because x and y are not elements of R². What are the domain and codomain of f? What do elements of the domain and codomain look like?

domain is x+y, codomain is y

 

[vela]

No. When you write f:A→B, that means A is the domain and B is the codomain. So in your problem, both are R². So f maps one ordered pair (x,y) in the domain to the ordered pair (x+y,y) in the codomain.

 

[kathrynag]

Ok so then I need to show f(a)=b
f(x,y)=(x+y,y)
f(a)=(x+y,y)

 

[kathrynag]

I was too sloppy. It's not all subsets of SxT but only those which satisfy the definition of a function, namely that it's single-valued.

Suppose you have a domain S={1,2,3} and codomain T={a,b,c}. The cartesian product SxT is the set consisting of all ordered pairs: SxT={(x,y) | x is in S and y is in T}. In this case, you'd have

SxT={(1,a), (1,b), (1,c), (2,a), (2,b), (2,c), (3,a), (3,b), (3,c)}

A function f from S to T is a subset of SxT which satisfies the requirement that an element in S maps to one value in T. For example, if you had

f={(1,a), (2,b), (3,c)}

this would be the function where, in more common notation, f(1)=a, f(2)=b, and f(3)=c. On the other hand, the subset

g={(1,a), (1,b)}

is not a function because it would mean that g(1)=a and g(1)=b, which you know isn't true for a function.

I don't understand why the codomain is not {4,5}

 

[kathrynag]

I was too sloppy. It's not all subsets of SxT but only those which satisfy the definition of a function, namely that it's single-valued.

Suppose you have a domain S={1,2,3} and codomain T={a,b,c}. The cartesian product SxT is the set consisting of all ordered pairs: SxT={(x,y) | x is in S and y is in T}. In this case, you'd have

SxT={(1,a), (1,b), (1,c), (2,a), (2,b), (2,c), (3,a), (3,b), (3,c)}

A function f from S to T is a subset of SxT which satisfies the requirement that an element in S maps to one value in T. For example, if you had

f={(1,a), (2,b), (3,c)}

this would be the function where, in more common notation, f(1)=a, f(2)=b, and f(3)=c. On the other hand, the subset

g={(1,a), (1,b)}

is not a function because it would mean that g(1)=a and g(1)=b, which you know isn't true for a function.

Could I say we have a function defined by f(1)=4, f(2)=4, f(3)=4
Another function defined by f(1)=5, f(2)=5, f(3)=5
Another defined by f(1)=4, f(2)=4, f(3)=5
And so on... Is that the idea?

 

[vela]

I don't understand why the codomain is not {4,5}

Oh, I was just giving you an example. The codomain is {4,5} in your problem.

Could I say we have a function defined by f(1)=4, f(2)=4, f(3)=4
Another function defined by f(1)=5, f(2)=5, f(3)=5
Another defined by f(1)=4, f(2)=4, f(3)=5
And so on... Is that the idea?

Exactly.

 

[kathrynag]

Ok so then I need to show f(a)=b
f(x,y)=(x+y,y)
f(a)=(x+y,y)

Or could I say something like f(x0,y0)=(x,y) is what I want to show.
Then f(x0,y0)=(x0+y0,y0)
So not onto.