List the things f has to do to be a homomorphism

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In summary, a Homomorphism of groups must send the identity element of the domain to the identity element of the codomain, and a function must be defined on all elements of the given domain.
  • #1
Bachelier
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The function

##f(x) = \frac{x}{x+1} ##

is not a Homomorphism because f(1) ≠ 1..Am I correct?
 
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  • #2


List the things f has to do to be a homomorphism.
 
  • #3


Simon Bridge said:
List the things f has to do to be a homomorphism.

Assume we are in ℝ then

f(ab) = f(a).f(b) under multiplication
f(a+b) = f(a)+f(b) under addition

and that's all I find in my book. but I know we need to check some extra stuff.
 
  • #4


OK - but if it fails anyone of the conditions, then it isn't a homomorphism right?
Have you applied either of those two tests to this situation?

Your example was f(1)=1 - proposed as a test.
How does that work in with the relations you listed?
In the first, ab=1 and in the second a+b=1.

You could also consider what sort of transformation is represented by f(x) ... i.e. is f(x) defined for all real x? Does it have to be if it is to be a homomorphism?
 
  • #5


Are you sure that you don't mean homeomorphism, not homomorphism?
 
  • #6


I'll assume you mean homeomorphism then f(x) must be continuous and have a continuous inverse f(x) is not continuous at x = -1 and the inverse function is not continuous at f^-1=1.
 
  • #7


Bachelier said:
Assume we are in ℝ then

f(ab) = f(a).f(b) under multiplication
f(a+b) = f(a)+f(b) under addition

and that's all I find in my book. but I know we need to check some extra stuff.

Could you please always list what structure you are working with. Saying that "f is a homomorphism" is a meaningless statement. You should state "f is a homomorphism of groups/rings/fields/algebras/lattices/..."

Also, be sure to always give the domain and codomain.
 
  • #8


micromass said:
Could you please always list what structure you are working with. Saying that "f is a homomorphism" is a meaningless statement. You should state "f is a homomorphism of groups/rings/fields/algebras/lattices/..."

Also, be sure to always give the domain and codomain.

Homomorphism of groups. mainly define our f: [0,∞) → ℝ.

I swear sometimes I just need some sleep. of course it is not because it fails property 1. mainly f(a+b) ≠ f(a) + f(b).
Sometimes my brain will jump to the most complex property and try to solve it while ignoring the simplest ones.

That aside, a Homomorphism of groups must send the identity element of the domain to the identity element of the codomain, right?
 
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  • #9


Simon Bridge said:
You could also consider what sort of transformation is represented by f(x) ... i.e. is f(x) defined for all real x? Does it have to be if it is to be a homomorphism?

If I define my function f from ℝ to ℝ then the function is not defined at x = -1.

I take it from your question that a Homomorphism of groups must be well-defined on all elements of the group?
 
  • #10


Bachelier said:
Homomorphism of groups. mainly define our f: [0,∞) → ℝ.

OK, but [itex][0,\infty)[/itex] is not a group. So you can't talk about homomorphism of groups. Furthermore, a group only has one operation. So, saying that a homomorphisms of groups satisfy

[tex]f(x+y)=f(x)+f(y)~\text{and}~f(xy)=f(x)f(y)[/tex]

is not correct. Why not? Because now you're talking about two operations: addition and multiplication. A group is a set with only one operation (which satisfies some conditions.

So if you have a function [itex]f:(\mathbb{R},+)\rightarrow (\mathbb{R},+)[/itex] (I usually denote a group by [itex](G,*)[/itex], where G is a set and * is an operation on the set), then this is a homomorphism if and only if [itex]f(x+y)=f(x)+f(y)[/itex]. The multiplication has nothing to do with this.

In general, a function [itex]f:(G,*)\rightarrow (H,\oplus)[/itex] must satisfy [itex]f(x*y)=f(x)\oplus f(y)[/itex]. Nothing more.

If you want to talk about two operations (like addition and multiplication on [itex]\mathbb{R}[/itex]), then you have to talk about rings.

That aside, a Homomorphism of groups must send the identity element of the domain to the identity element of the codomain, right?

Yes. But what you mean with identity element depends on the group operation. In the group [itex](\mathbb{R},+)[/itex], the identity is 0. In the groups [itex](\mathbb{R}\setminus\{0\},\cdot)[/itex], the identity is 1.
 
  • #11


Bachelier said:
If I define my function f from ℝ to ℝ then the function is not defined at x = -1.

I take it from your question that a Homomorphism of groups must be well-defined on all elements of the group?

Yes. But that's not only true for homomorphisms. It is in general true for functions. A function [itex]f:X\rightarrow Y[/itex] must be defined on all [itex]x\in X[/itex].

So [itex]f:\mathbb{R}\rightarrow \mathbb{R}:x\rightarrow \frac{1}{x}[/itex] is not a function (because not defined in 0). But [itex]f:\mathbb{R}\setminus \{0\}\rightarrow \mathbb{R}:x\rightarrow \frac{1}{x}[/itex] is a function.
 
  • #12


micromass said:
OK, but [itex][0,\infty)[/itex] is not a group. So you can't talk about homomorphism of groups. Furthermore, a group only has one operation. So, saying that a homomorphisms of groups satisfy

[tex]f(x+y)=f(x)+f(y)~\text{and}~f(xy)=f(x)f(y)[/tex]

is not correct. Why not? Because now you're talking about two operations: addition and multiplication. A group is a set with only one operation (which satisfies some conditions.

So if you have a function [itex]f:(\mathbb{R},+)\rightarrow (\mathbb{R},+)[/itex] (I usually denote a group by [itex](G,*)[/itex], where G is a set and * is an operation on the set), then this is a homomorphism if and only if [itex]f(x+y)=f(x)+f(y)[/itex]. The multiplication has nothing to do with this.

In general, a function [itex]f:(G,*)\rightarrow (H,\oplus)[/itex] must satisfy [itex]f(x*y)=f(x)\oplus f(y)[/itex]. Nothing more.

If you want to talk about two operations (like addition and multiplication on [itex]\mathbb{R}[/itex]), then you have to talk about rings.



Yes. But what you mean with identity element depends on the group operation. In the group [itex](\mathbb{R},+)[/itex], the identity is 0. In the groups [itex](\mathbb{R}\setminus\{0\},\cdot)[/itex], the identity is 1.

Beautiful Math and very clear definitions..Thank you very much.

[0,∞) is indeed not a group because of the inverse axiom. For instance wrt multiplication 0 has no inverse. (if the operation is addition, then no element has an inverse)

So I guess I can only call [0,∞) an interval or a set.

Now ##((0,∞), *)## where * is the regular multiplication is a group under ##*##, right?
 
  • #13


Bachelier said:
Now ##((0,∞), *)## where * is the regular multiplication is a group under ##*##, right?

Yes, it is. You may be surprised to learn that [itex]((0,+\infty),\cdot)[/itex] is actually isomorphic (as group) to [itex](\mathbb{R},+)[/itex].

Indeed, the isomorphism is

[tex]T:(\mathbb{R},+)\rightarrow ((0,+\infty),\cdot):x\rightarrow e^x.[/tex]
 
  • #14


micromass said:
Yes, it is. You may be surprised to learn that [itex]((0,+\infty),\cdot)[/itex] is actually isomorphic (as group) to [itex](\mathbb{R},+)[/itex].

After learning a long time ago that ##(0,1) \cong \mathbb{R}##, nothing surprises me anymore... lol :)
 
  • #15


Bachelier said:
After learning a long time ago that ##(0,1) \cong \mathbb{R}##, nothing surprises me anymore... lol :)

OK, but that's not as groups.
 

What is a homomorphism?

A homomorphism is a mathematical function that preserves the structure of a mathematical object. In simpler terms, it is a function that maps one algebraic structure to another in a way that respects their operations and properties.

What does it mean for a function to be a homomorphism?

For a function to be a homomorphism, it must preserve the operations and properties of the mathematical objects it is mapping. This means that the function must produce the same result as the original objects when the operations are performed on them.

What are the things f has to do to be a homomorphism?

To be a homomorphism, the function f must satisfy the following conditions:

  • Preserve the operation: f(a * b) = f(a) * f(b) for all elements a and b in the domain.
  • Preserve the identity element: f(e) = e', where e is the identity element in the domain and e' is the identity element in the codomain.
  • Preserve inverses: f(a^-1) = (f(a))^-1, where a^-1 is the inverse of element a in the domain and (f(a))^-1 is the inverse of f(a) in the codomain.

What is the difference between a homomorphism and an isomorphism?

While both homomorphisms and isomorphisms preserve the structure of mathematical objects, the main difference is that isomorphisms are bijective, meaning they have a one-to-one correspondence between the elements of the domain and codomain. Homomorphisms, on the other hand, do not necessarily have a one-to-one correspondence between the elements.

How are homomorphisms used in science?

Homomorphisms are used in various fields of science, such as physics, chemistry, and computer science. They provide a way to study and understand the relationships between different mathematical objects and their structures. In physics, homomorphisms are used to describe the symmetries of physical systems, while in chemistry, they help in understanding the properties of molecules. In computer science, homomorphisms are used in data encryption and coding theory.

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