List the things f has to do to be a homomorphism

[Bachelier]

The function

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

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

 

[Simon Bridge]

List the things f has to do to be a homomorphism.

 

[Bachelier]

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.

 

[Simon Bridge]

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?

 

[Jim Kata]

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

 

[Jim Kata]

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.

 

[micromass]

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.

 

[Bachelier]

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?

 

[Bachelier]

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?

 

[micromass]

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

OK, but [0,\infty) 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

f(x+y)=f(x)+f(y)~\text{and}~f(xy)=f(x)f(y)

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

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

If you want to talk about two operations (like addition and multiplication on \mathbb{R}), 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 (\mathbb{R},+), the identity is 0. In the groups (\mathbb{R}\setminus\{0\},\cdot), the identity is 1.

 

[micromass]

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 f:X\rightarrow Y must be defined on all x\in X.

So f:\mathbb{R}\rightarrow \mathbb{R}:x\rightarrow \frac{1}{x} is not a function (because not defined in 0). But f:\mathbb{R}\setminus \{0\}\rightarrow \mathbb{R}:x\rightarrow \frac{1}{x} is a function.

 

[Bachelier]

OK, but [0,\infty) 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

f(x+y)=f(x)+f(y)~\text{and}~f(xy)=f(x)f(y)

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

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

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



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

 

[micromass]

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

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

Indeed, the isomorphism is

T:(\mathbb{R},+)\rightarrow ((0,+\infty),\cdot):x\rightarrow e^x.

 

[Bachelier]

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

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

 

[micromass]

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

OK, but that's not as groups.