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Let $R$ an equivalence relation over a set $A$. For $a \in A$ we define $a / R = \{ b \in A: (a, b) \in R\}$ and call this set the equivalence class of $A$. Furthermore, we define $A / R = \{a / R : a \in A\}$ the set of all equivalence classes for the equivalence relation $R$ over $A$. As a last definition, if $f : A \mapsto B$ we define $ker ~ f = \{(a, b) \in A^2 : f(a) = f(b)\}$.

In general, a function $f : A / R \mapsto B$ is ambiguous or contradictory. For example, if we define $f(a / R) = a^2$ and $R$ is the relationship "has the same parity", then the fact that $2 / R = 4 / R$ would lead us to expect $f(2 / R) = 4 = f(4 / R) = 16$.

Notwithstanding, an important idea in modern algebra is the following:

If $f : A \mapsto B$ is onto, then $\overline{f}(a / ker ~ f) = f(a)$ defines a bijection $\overline{f} : A / ker ~ f \mapsto B$.

It is clear to me $a.$ that $\overline{f}$ is a well-defined function, in the sense that to each element in the domain it maps a unique element in the codomain, and $b.$ that it is a bijection. However, I do not see why we need to assumie that $f : A \mapsto B$ is onto (or surjective) in order for the statement to be true. To prove that it is a function and that it is a bijection we could simply make the following observations:

Because by definition $f(a) = f(b)$ for any $(a, b) \in ker(f)$, it is clear that $\overline{f}(a / ker ~ f) = f(a)$ is a function. That it is a bijection follows from the following: if $f(a) = \overline{f}(a / ker ~ f)$ and $f(a) = \overline{f}(b / ker ~ f)$ then, by definition, $(a, b) \in ker(f)$, which implies $a / ker ~ f = b / ker ~ f$. Then $\overline{f}$ is injective. Furthermore, any $b \in B$ is at least equivalent to itself under any equivalence relation, including $ker ~ f$, and hence belongs to the equivalence class $b / ker ~ f$. This should suffice to show $\overline{f}$ is surjective.

As you can see, in no point did we need to assume the surjectivity of $f$ to adscribe to $\overline{f}$ the properties in question. I must be missing something, so the question is: why do we need to assume $f$ is surjective in order for $\overline{f}$ to satisfy the statement?

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  • $\begingroup$ Hello. I don't really see how your argument shows $f$ is a bijection. You seem to be showing it's injective? Are you aware that not every injection is a bijection? $\endgroup$ Mar 2 at 14:47
  • $\begingroup$ Thanks for pointing that out, I wrote that parte of the question in a hurry. I've edited it $\endgroup$
    – lafinur
    Mar 2 at 15:48
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    $\begingroup$ JonathanZ correctly points out your mistake. In situations like this it can be useful to look at a concrete example to see what goes wrong in your argument. (For example think about the function $\Bbb Z \to \Bbb Z$ sending $n \mapsto 2n$, or the one sending $n \mapsto 0$). You might like to try to show that $\overline f: A / \ker f \to B$ is a bijection if and only if $f$ is surjective, indicating that "$f$ is surjective" is necessary in the strongest possible sense. $\endgroup$ Mar 2 at 16:06
  • $\begingroup$ Another way to see if what you're proposing is sensible is to think about what would happen if conditions were varied a bit. In this case, suppose in your version you already had your bijection $A/ker f \rightarrow B$. If someone came along and added a few new elements to $B$, you'd still satisfy the requirements of your version , but that map would certainly no longer be a surjection. Something has to be wrong then. $\endgroup$
    – JonathanZ
    Mar 2 at 16:18

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In your proof of surjectivity, you've taken an element $b\in B$, but then started talking about $b/ ker f$. That's wrong - $x/ker f$ only makes sense for $x$ in $A$, and elements of $B$ do not have to be elements of $A$.

The more general lesson we can learn here is that when you've got sets named $A$ and $B$ laying around, notation like $ker ~ f = \{(a, b) \in A^2 : f(a) = f(b)\}$ is risky - it can easily lead you into mistakes like this. I'd recommend $(a,a')$ or $(a_1, a_2)$ for a generic element of $A\times A$ - let the notation help you, not trick you.

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    $\begingroup$ Thanks for answering the question and also the more general advice. :) $\endgroup$
    – lafinur
    Mar 3 at 20:24

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