
#1
Feb1513, 07:25 PM

P: 376

The function
##f(x) = \frac{x}{x+1} ## is not a Homomorphism because f(1) ≠ 1..Am I correct? 



#2
Feb1513, 10:21 PM

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




#3
Feb1513, 11:07 PM

P: 376

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
Feb1513, 11:19 PM

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List the things f has to do to be a homomorphism
OK  but if it fails any one 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? 



#6
Feb1613, 02:07 AM

P: 233

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
Feb1613, 02:23 AM

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



#8
Feb1613, 07:05 AM

P: 376

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? 



#9
Feb1613, 07:14 AM

P: 376

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



#10
Feb1613, 07:36 AM

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[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. 



#11
Feb1613, 07:38 AM

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P: 16,591

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
Feb1613, 09:09 PM

P: 376

[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
Feb1613, 09:12 PM

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P: 16,591

Indeed, the isomorphism is [tex]T:(\mathbb{R},+)\rightarrow ((0,+\infty),\cdot):x\rightarrow e^x.[/tex] 



#14
Feb1613, 10:32 PM

P: 376




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