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I have a question which is as follows,

Consider $f:\mathbb{R} \to \mathbb{R}$ and $g:\mathbb{R} \to \mathbb{R}$. Define the sum of $f$ and $g$ as the funtion $h:\mathbb{R} \to \mathbb{R}$ such that $h(x)=f(x)+g(x)$ for all $x \in \mathbb{R}$.

If both $f$ and $g$ are injective, will $h$ be injective? If only one of $f$ or $g$ is injective will $h$ be injective?

And also, if both $f$ and $g$ are surjective, will $h$ be surjective? If only one of $f$ or $g$ is surjective will $h$ be surjective?

I think that if both $f$ and $g$ are injective, then $h$ may not be injective and I can find a counter example to show that when both $f$ and $g$ are injective, $h$ is not injective.

But I have no idea on how to approach the rest of the question. Please give me some hints on how to do this, especially the part of the question asking if only one of $f$ and $g$ is injective and surjective, thanks to anybody who can help.

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    $\begingroup$ Take one bijection $f$ and then another bijection $g = -f$ and $f + g$ is zero every time. $\endgroup$
    – brick
    Oct 28, 2014 at 20:02

2 Answers 2

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Hint: Take $f(x)=x$ and $g(x)=-x$. Works as a counter-example for both.

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  • $\begingroup$ Thanks, but what should I do with the part of the question asking only one of f or g is injective or surjective? $\endgroup$
    – Lucy
    Oct 28, 2014 at 20:05
  • $\begingroup$ But isn't both $f(x)=e^x$ and $g(x)=e^{-x}$ injective? $\endgroup$
    – Lucy
    Oct 28, 2014 at 20:17
  • $\begingroup$ Oh sorry. I'll think over it. $\endgroup$ Oct 28, 2014 at 20:20
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Both are false- and "one example rules them all".

Take $f$ to be any bijective function, and $g(x)=-f(x)$.

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  • $\begingroup$ Thank you, but what about when only one of $f$ and $g$ is injective or surjective? Are both still false? $\endgroup$
    – Lucy
    Oct 28, 2014 at 20:19
  • $\begingroup$ @Lucy: yes.. just take $g=-f$ $\endgroup$
    – voldemort
    Oct 28, 2014 at 21:45

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