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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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What is the map $f$ doing? Do you mean to say "$S$ and $T$ are isomorphic under $f$"? If so, $f$ must be 1-1, as this is part of the definition of isomorphism. – Austin Mohr Sep 19 '11 at 3:41
Let $S$ and $T$ be two one-element sets. Then there is a unique function $S \to T$ and it is bijective. Conversely, if $S$ and $T$ are in bijection but there is a map $S \to T$ which is not bijective, then $T$ must contain at least two elements. – Zhen Lin Sep 19 '11 at 3:44
Any old $f$ between two isomorphic sets need not be a witness to the fact that $S$ and $T$ are isomorphic, so it does not have to be 1-to-1. E.g., $S=\{a,b\}$, $T=\{1,2\}$ are bijectable/isomorphic as sets, but $f\colon S\to T$ given by $f(a)=f(b)=1$ is a perfectly fine function that is not 1-to-1. – Arturo Magidin Sep 19 '11 at 3:49
To combine Zhen and Arturo's answers: a constant function on a set of more than one element will do the trick. – Adam Saltz Sep 19 '11 at 3:50
@IAmBrianDawkins It seems Jeff wanted a function that is onto but not one-to-one, in which case Zhen and Arturo's answers apply only to sets of 2 elements. The function $h:\mathbb{R} \to \mathbb{R}$ given by $h=c$ is certainly not onto. – barf Oct 20 '11 at 7:48

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., $f(8) = f(9) = 4$.)

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