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I can't seem to wrap my head around writing a function as the composition of two other functions under the constraint that one of the functions must be injective and the other must be surjective. $ f:\mathbb{R} \to \mathbb{R}$

I am trying to write: \begin{align} f(x) = |x| + 1\\ \end{align} I know \begin{align} \sqrt[]{x^2} \end{align} is equivalent to |x|, but this function is not injective or surjective, so I believe my best option is to add a term of x to this to make it one of either, for example adding x for injective or x^3 for surjective. I have no idea where to go from here.

I am really looking for general advice so I can tackle these problems as they come.

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Any information about what the domain and range have to be? or do both have to be from R->R? –  Sun Oct 10 '13 at 14:15
    
Yes sorry, both from R->R. I will edit that in. @Sun –  Display Name Oct 10 '13 at 14:17
    
No what I meant was, you want to write $f$ as $g \circ h$, right? Are both $g$ and $h$ defined as surjective or injective on $\mathbb{R} \rightarrow \mathbb{R}$ or can we choose the sets where they are defined? –  Sun Oct 10 '13 at 14:24
    
You have to define the set you're targeting when defining a surjective function. Surjective on what? On $\mathbb{R}$ or on $[1,\infty)$ ? And does the order matter? There is no surjective function to $\mathbb{R}$ which only has values from the set $[1,\infty)$ you know. –  Zafer Sernikli Oct 10 '13 at 14:26
    
Yes f as g(h(x)) with g and h both on $\mathbb{R}$→$\mathbb{R}$ with one being surjective and one being injective.Sorry for ambiguity!! @Sun –  Display Name Oct 10 '13 at 14:28

1 Answer 1

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Consider a composition $f = h \circ g$. Also, there exists a bijection $k$ between $\mathbb{R}^{+} \rightarrow \mathbb{R}$ - there are many ways to construct one. One example is to make $[0 \quad 1]$ map to $[-1 \quad 1]$ and similarly extend. Let $g = k(|x|)$. This is a surjection. Now, $h = k^{-1} + 1$. It is defined from $\mathbb{R} \rightarrow [1,\infty)$ and bijective there, but when the range is embedded within $\mathbb{R}$ , $h$ is injective.

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I think I may be too tired to understand, but for f(x) = h(g(x)), I have nothing for h to take in in terms of x values, so h isn't referenced. If I take it as h(g(k)), I get h(k(abs(x))) which gives me (k(abs(x)))^-1 + 1, which I don't believe equals abs(x) + 1. –  Display Name Oct 10 '13 at 19:50
    
k^-1(k(x)) = x. So k^-1(k(|x|)) = abs(x) - I think I get it now, I had a quick nap. –  Display Name Oct 10 '13 at 21:05

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