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I am struggling with getting an example of two sets S and T and (onto) mapping f, where the fact S and T are isomorphic does not imply that f is also 1 - 1. If possible could you also give an example in which the fact that they are isomorphic would imply that they are 1 - 1?

Thank You!

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    @Iambriandawkins Ah, I will be sure to check that next time before 'correcting'!2011-10-21

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Hint: Can you do this for finite $T$ and $S$? What happens if $T$ and $S$ are infinite? Think of shift-like maps $\mathbb{N}\rightarrow \mathbb{N}$

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Consider the map $f: \mathbb{N} \to \mathbb{N}$ given by $\begin{aligned} 1 &\mapsto 1,\\ 2 &\mapsto 1,\\ 3 &\mapsto 2,\\ 4 &\mapsto 3, \\ 5 &\mapsto 4,\\ & \ \ \vdots\end{aligned}$ ie $f(n) = \begin{cases} 1 &\text{ if } n =1,\\ n - 1 &\text{ otherwise.} \end{cases} $

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The comments and answer so far given seem to ignore the requirement that $f$ be onto. Here’s an example that takes that into account. Take $S=T=\mathbb{N}$, and let $f:S\to T:n\mapsto \left\lfloor\frac{n}2\right\rfloor.$ You should be able without too much trouble to show that $f$ us surjective (maps $S$ onto $T$) and is two-to-one everywhere. (E.g., $f(8) = f(9) = 4$.)