I have a function $f$ that is injective from $\mathbb R \to \mathbb R$ but not surjective, i.e., the image $f$ is not all of $\mathbb R$. Can I make $f$ bijective by patching the codomain and removing any parts that are not in the image?
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1Yes, you can, but the new function is different from the older one, it is not the same function strictly speaking. – Masacroso Sep 05 '16 at 02:52
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Yes. Then you have a new function $g \colon \mathbb{R} \to \operatorname{Im} (f)$ such that $g(r) := f(r)$ for each $r \in \mathbb{R}$, which is pretty clearly surjective. This is called restricting the codomain. Note that the function doesn't need to be injective to do this construction.
Similarly, if your function isn't injective, you can restrict the domain to make an agreeing injective function. But there isn't a unique way to do this.
Mike Pierce
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1in order to the function to be bijective, it must be injective as well, is that right? – Rodrigo Stv Sep 05 '16 at 02:55
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@RodrigoStv, yeah it needs to be injective first in order to be a bijection after you restrict the codomain. – Mike Pierce Sep 05 '16 at 02:58