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$f$ and $g$ are bijection from $[0,1]$ to $[0,1]$. Which of the following are one to one (injection) from $[0,1]$ to $[0,1]?

  1. $f+g$
  2. $f-g$
  3. $f*g$
  4. $\frac{1}{2}(f^2+g^2)$
  5. $f \circ g$

I guess 1,2 and 5 are injection. Is that correct? Could anyone tell me how to quickly judge which one is injection?

Thanks : )

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You might first consider which of these are guaranteed to map into $[0,1]$. –  AMPerrine Oct 30 '12 at 17:08
    
Be careful. Suppose there are distinct $x,y \in [0,1]$ with $f(x) = 0$, $g(x) = 1$, $f(y) = g(y) = 1/2$. It follows $(f+g)(x) = (f+g)(y) = 1$, so the sum is not an injection. –  Austin Mohr Oct 30 '12 at 17:08
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2 Answers 2

up vote 4 down vote accepted

5 is an injection as the composition of two injections. All the rest are in general not injections. For counterexamples you can use $f(x)=x$ and $g(x)=1-x$ or, in case 2, $f(x)=g(x)=x$.

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It’s always a good idea to look at some simple examples to try to get a feel for what’s going on. In (2), for instance, what happens if $f=g$?

For that matter, what happens in (1)? Since $f=g$ is a bijection from $[0,1]$ to $[0,1]$, there must be some $x\in[0,1]$ such that $f(x)=g(x)=1$, so that $(f+g)(x)=2\notin[0,1]$. If that’s not good enough $-$ after all, $f+g$ might still be a bijection between $[0,1]$ and some set $-$ consider two of the simplest bijections between $[0,1]$ and itself: $f(x)=x$, and $g(x)=1-x$. With that pair you can settle both (1) and (3) immediately, and with a little more thought, perhaps aided by looking at their graphs, you can settle (4) as well.

That leaves only (5), and you should already know that the composition of bijections is a bijection.

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@Ross: You missed the hypothesis that $f=g$. –  Brian M. Scott Oct 30 '12 at 17:35
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