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Let $\alpha : F(\mathbb{R}) \to \mathbb{R}$ be defined by $\alpha(f)=f(1)$ and let $\beta : F(\mathbb{R}) \to \mathbb{R}$ be defined by $\beta(f)=f(2)$

Prove that $\alpha$ and $\beta$ are homomorphisms from $F(\mathbb{R})$ onto $\mathbb{R}$.

I'm having trouble with this question because it is a bit confusing to me. I know homomorphisms are defined by a function and a property. In this particular case, the function is known which is $\alpha$ and $\beta$ but showing the property of homomorphism is confusing. Do I treat each separately or as one?

Meaning, do I need to show:

$$\alpha(fg)=f(1)g(1) \text{ where } f,g \in F(\mathbb{R})$$

or do I need to show

$$\alpha(f)\beta(f)=f(1)f(2)?$$

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  • $\begingroup$ What exactly is $F(\Bbb R)$..? $\endgroup$ – Cameron Williams Apr 21 '15 at 0:54
  • $\begingroup$ $F(\mathbb{R})$ is the set of all functions each of which has a domain and range of $\mathbb{R}$. $\endgroup$ – hchar Sep 8 at 2:14
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The former. That is, you treat $\alpha$ and $\beta$ separately, they do not interact.

To show that $\alpha$ is a homomorphism, you need to show that $\alpha(fg) = \alpha(f)\alpha(g)$.

To show that $\beta$ is a homomorphism, you need to show that $\beta(fg) = \beta(f)\beta(g)$.

Think of this as a two part question where part one asks you to prove that $\alpha$ is a homomorphism and part two asks you to prove that $\beta$ is a homomorphism.

Added Later: I just noticed that you are not just asked to show that $\alpha$ and $\beta$ are homomorphisms from $F(\mathbb{R})$ to $\mathbb{R}$, but homomorphisms from $F(\mathbb{R})$ onto $\mathbb{R}$. So in addition to showing that $\alpha$ and $\beta$ are homomorphisms, you must also show that they are onto.

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  • $\begingroup$ The onto part is a bit tricky, what I am thinking is that if $f$ is in $F(\mathbb{R})$ then from above the output is $f(1)$ which is in the reals because any real times 1 will be a real. As for homomorphism, i did the following: for $f,g \in F(\mathbb{R}),$ we have $\alpha(fg)=fg(1)=f(1)g(1)=\alpha(f)\alpha(g)$ and it is similar for $\beta$ $\endgroup$ – Emmie Apr 20 '15 at 23:49
  • $\begingroup$ @Emmie: I'm a little bit confused by this: "$f(1)$ which is in the reals because any real times 1 will be a real." There is no multiplication going on so I don't know what you mean by "any real times 1". Note that $f(1)$ is not multiplication but function evaluation. $\endgroup$ – Michael Albanese Apr 20 '15 at 23:52
  • $\begingroup$ Oh sorry, for some reason I'm always confusing operations. To show onto, you pick an element in the range, in this case, it would be $\mathbb{R}$. So, I pick $f(1)=\alpha(f)=f(1)$ which shows $\alpha$ is onto? $\endgroup$ – Emmie Apr 27 '15 at 3:47
  • $\begingroup$ To show $\alpha$ is onto, you need to show that for any $c \in \mathbb{R}$, there is $f \in F(\mathbb{R})$ such that $\alpha(f) = c$. As $\alpha(f) = f(1)$, this means that you have to show that for any real number $c \in \mathbb{R}$, there is a function $f \in F(\mathbb{R})$ such that $f(1) = c$; i.e. there is a function $f \in F(\mathbb{R})$ that takes the value $c$ at $1$. $\endgroup$ – Michael Albanese Apr 27 '15 at 3:56
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Imagine functions as a huge object that nobody can see them in full. And $\alpha$ is a watchman at $1$, ( $\beta$ a watch man at 2 etc). You ask these guys any info about functions they can only give the value of the function at $1$. (and at 2 for $\beta$).

Now if you know $h=fg$, (otherwise nothing about the functions) and $\alpha$ gives you the values for $f$ and $g$ cannot you predict what $\alpha$ would say about $h$? (and similarly for $k=\alpha+\beta$?).

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