Consider the injection $f:{{0,1,2}} \to {0,1,2,3}:x\to x $
Now consider the left inverse of this function. How exactly is this a surjection if not all the elements in the codomain are mapped back?
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2$\begingroup$ This function has $3$ left inverses. Can you find them? $\endgroup$– BerciCommented Jan 10, 2021 at 18:17
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$\begingroup$ Ah so you're saying that there exists a surjective inverse? $\endgroup$– expl0it3rCommented Jan 10, 2021 at 18:19
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$\begingroup$ No, I'm asking you to explicitly write down the left inverses. Implicitly yes, I'm saying it has. $\endgroup$– BerciCommented Jan 10, 2021 at 18:21
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1$\begingroup$ No, how it could have? Specify your concern. Are you sure you are thinking about a function $\{0,1,2,3\}\to\{0,1,2\}$? $\endgroup$– BerciCommented Jan 10, 2021 at 18:27
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1$\begingroup$ if you want maps to be total, your diagram does not represent a left inverse. $\endgroup$– AurelioCommented Jan 10, 2021 at 18:28
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1 Answer
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The given function $f$ has exactly 3 left inverses.
Note that a left inverse $g$ of $f$ must be a function $\{0,1,2,3\}\to\{0,1,2\}$ so that $g(f(x))=x$ for all $x\in\{0,1,2\}$.
So, we must have $g(0)=0,\ g(1)=1,\ g(2)=2$, and also $g(3)\in\{0,1,2\}$, which actually can be arbitrary.
Because of the equation $g(f(x))=x$ we see that indeed any such $g$ must be surjective, since for arbitrary $x$ in the codomain of $g$ (=domain of $f$) we have an element, specifically e.g. $f(x)$ which $g$ carries into $x$.
