Let $f:X\to Y$ be a surjection. Let $R$ be the subset of $X\times X$ consisting of those pairs $(x,x')$ such that $f(x)=f(x')$. Then $R$ is an equivalence relation. Let $\pi:X\to X/R$ be the projection. Verify that, if $\alpha\in X/R$ is an equivalence class, to define $F(\alpha)=f(a)$, whenever $\alpha=\pi(a)$, establishes a well-defined function $F:X/R\to Y$ which is bijective.

Note: Let $E$ be a set and $R$ an equivalence relation. Then $E/R$ is the set of equivalence classes and $\pi:E\to E/R$ is surjective. $\pi$ is called the projection.

I'm not necessarily looking for the answer, but rather some insight in to how to approach the problem (or even better, a hint).

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  • You haven't made a clear question. Where are you having trouble? What are you not understanding? – Pedro Dec 23 '14 at 01:53
  • @PedroTamaroff I have to verify that $F$ is a well-defined bijection, but I'm not sure how to do that. – Mr.Young Dec 23 '14 at 01:55
  • Do you know what a bijection is? Do you know what it means for $F$ to be well-defined? Since $f$ is a surjection, and since $F$ is defined on classes $[a]$ as $F([a])=f(a)$, it should be clear $F$ is still surjective. So it suffices you see it is injective. – Pedro Dec 23 '14 at 01:58
  • @PedroTamaroff Sorry for not being more clear. I do understand that all I need to do is show that $F$ is injective, and I see that $F$ must be $f\circ\pi^{-1}$ some how. I'm just stuck on how to proceed. – Mr.Young Dec 23 '14 at 02:02

1 Answers1


The intuition here is that we have a surjection $f:X\to Y$ and we want to upgrade it to an injection by somehow modifying the domain $X$ into a new domain $\tilde{X}$ and procuring a better version, call it $\tilde{f}$, of $f$ from this new domain to $Y$.

A simple example is given by the map $f:\{1,2,3,4,5,6\}\to\{1,2,3\}$ defined by $$f(1)=1\\ f(2)=2\\ f(3)=2\\ f(4)=1\\ f(5)=3\\ f(6)=1$$

Our toy function $f$ fails to be injective because it sends $1$, $4$, and $6$ all to $1$, and $2$ and $3$ both to $2$. One way to "injectify" $f$ would be to pick a subset of the domain, like $\tilde{X} = \{1,2,5\}$, and let $\tilde{f}$ be the restriction of $f$ to $\tilde{X}$. Sure, $\tilde{f}$ is now injective, but why did we pick the set $\{1,2,5\}$? Why not, say, $\{3,4,5\}$ or $\{3,5,6\}$? None of these choices is natural.

We remedy this by bunching together all the points where $f$ fails to be injective. Define a new domain $\tilde{X} = \{\{1,4,6\}, \{2,3\}, \{5\}\}$ whose elements are sets of numbers, as opposed to just numbers. (This shouldn't be an intuitive obstacle, because, at the end of the day, everything is a set.) Note, in particular, that $5$ gets put in a box all by itself, and appears within $\tilde{X}$ as $\{5\}$, not as $5$. This is so that the elements of $\tilde{X}$ have a guaranteed uniform type: they are subsets of $X$, not a mixture of subsets and elements of $X$.

How should we define $\tilde{f}$ on $\tilde{X}$? Let's go step by step. Where should $\tilde{f}$ send $\{1,4,6\}$? The obvious choice is $1$, because $1$ is equal to $f(1)$ and to $f(4)$ and to $f(6)$. The set $\{1,4,6\}$ was defined entirely in terms of $f$ (indeed, $\{1,4,6\}$ is precisely the preimage of $1$ under $f$) so it's most natural to define $\tilde{f}(\{1,4,6\})$ also entirely in terms of $f$. Making similar definitions for $\{2,3\}$ and $\{5\}$, we see that the map $\tilde{f}:\tilde{X}\to Y$ looks like this: $$\tilde{f}(\{1,4,6\}) = 1\\ \tilde{f}(\{2,3\}) = 2\\ \tilde{f}(\{5\}) = 3$$ We win! $\tilde{f}$ is injective, and surjective, hence bijective.

So what did we do, really? A typical element of $\tilde{X}$ is a biggest-possible set of points of $X$ that $f$ cannot tell apart. Your equivalence relation is $x\sim x'$ iff $f(x) = f(x')$. Personally, I call this "equality in the eyes of $f$", and it is a very common source of equivalence relations. The relation itself as a subset of $X\times X$ is called the kernel of $f$; you can read about that here: http://en.wikipedia.org/wiki/Kernel_%28set_theory%29. Note that my $\tilde{X}$ is equal to your $X/R$.

Finally, a note on the projection map $\pi$. In our toy example, there is a pretty obvious map from $\{1,2,3,4,5,6\}$ to $\{\{1,4,6\},\{2,3\},\{5\}\}$: send a number to the box containing it. That's all that $\pi$ is doing.

Armed with this intuition, I hope you can work out the details of the general case!

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