# Relationship between the quotient space of an equivalence relation and the kernel of a function

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?

• 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? Mar 2 at 14:47
• Thanks for pointing that out, I wrote that parte of the question in a hurry. I've edited it Mar 2 at 15:48
• 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. Mar 2 at 16:06
• 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. Mar 2 at 16:18

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.