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$f: \mathbb{R^+} \rightarrow \mathbb{R}: f(x) = \log_{10}(x)$

How is this function onto? I know it is one-to-one because every image has a pre-image. Can someone explain to me why this is also onto? I thought that everything in the codomain must be mapped to to be onto?

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    $\begingroup$ "one-to-one" does not mean that every image has a preimage. There needs to be uniqueness there. $\endgroup$
    – Dave
    Dec 15, 2020 at 21:09

2 Answers 2

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$f$ is onto $\mathbb R$, because, for any $y\in \mathbb R$, $f$ maps $10^y$ to $y$.

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Actually, $g\colon \Bbb R\to \Bbb R^+$, $x\mapsto 10^x$ turns out ot be a two-sided inverse of $f$, i.e.,

  • $g\circ f\colon \Bbb R^+\to \Bbb R^+$, $x\mapsto 10^{\log_{10}x}$ is the identity map of $\Bbb R^+$
  • $f\circ g\colon \Bbb R\to \Bbb R$, $x\mapsto \log_{10}(10^x)$ is the identity map of $\Bbb R$

Having a two-sided inverse is equivalent to being injective.

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  • $\begingroup$ I think you meant to write "equivalent to being bijective" (injective is equivalent to having a left inverse). $\endgroup$
    – mrtaurho
    Dec 16, 2020 at 23:15

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