# Contradiction?: Partitions/Equivalence classes vs. Bijective homomorphism

I am reading through Michael Artin's Algebra, and the following passages seem contradictory to me.

On page 55: "The elements of the image correspond bijectively to the nonempty fibres, which are the equivalence classes."

On page 56: Proposition 2.7.15: "Let K be the kernel homomorphism $\phi: G \rightarrow G'$. The fibre of $\phi$ that contains an element $a$ of $G$ is the coset $aK$ of $K$. These cosets partition the group $G$, and they correspond to elements of the image of $\phi$."

I would like to know what else is in the equivalence class of a? It could be alone, but it could also be b, and b~a, so $\phi(b) = \phi(a)$. However, if the partition of $aK$ all equal $\phi(a)$, then how can $\phi$ be a bijective map? Is it because everything in $aK$ is just the same thing? If this is the case, then groups are sets, and I was under the impression that in sets, repetition of elements is not allowed.

If anyone could clear my confusion, that would be very much appreciated. If there are simple examples that demonstrate what the textbook is trying to say, that would also be very helpful.

Many thanks.

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Who said $\,\phi\,$ is, or has to be, bijective?! All the elements in $\,aK=a\ker \phi\,$ are mapped to the same element by $\,\phi\,$ . And what do you mean by "if this is the case then groups are sets"? Of course they're sets: that's the very first, basic axiom of group theory! BTW, they're non-empty sets. – DonAntonio Aug 25 '12 at 22:59
It says from the excerpt on page 55 – fourdriver01 Aug 25 '12 at 23:25
No, it doesn't. It says there that $\,\overline\phi\,$ is a bijection, and this between fibers and the image of $\,\phi\,$ . – DonAntonio Aug 25 '12 at 23:33
I'm sorry I don't quite understand. Could you explain? Thanks – fourdriver01 Aug 26 '12 at 0:22
I added an answer (as it was too long for a comment) to address what I think could be some points you might be having trouble with. Read it down here. – DonAntonio Aug 26 '12 at 2:15

If $f:X\to Y$ is any map, the fibres of $f$ are the sets $f^{-1}[\{y\}]$ such that $y\in Y$; the particular set $f^{-1}[\{y\}]$ is called the fibre over $y$. The set $$\mathscr{F}=\Big\{f^{-1}[\{y\}]:y\in f[X]\Big\}$$ of all fibres of $f$ is a partition of $X$; it corresponds to the equivalence relation $x_1\sim x_2$ iff $f(x_1)=f(x_2)$. The correspondence $$y\leftrightarrow f^{-1}[\{y\}]\tag{1}$$ between $f[X]$ and $\mathscr{F}$ is clearly a bijection, and this has nothing to do with any extra structure on $X$ or $Y$. (It may not be a bijection between all of $Y$ and $\mathscr{F}$, because $f$ may not be a surjection: if $y\in Y\setminus f[X]$, then the fibre over $y$ is empty.) Note that the correspondence $(1)$ is not a bijection between $X$ and $f[X]$: it’s a bijection between $\mathscr{F}$, the set of non-empty fibres of $f$, and $f[X]$.

In the case of a group homomorphism $\varphi:G\to G'$, if $K=\ker\varphi$, then for $a,b\in G$ we have $\varphi(a)=\varphi(b)$ iff $a\in bK$ iff $b\in aK$. That is, $a$ and $b$ are in the same fibre of $\varphi$ iff they are in the same coset of $K$ in $G$. Thus, the non-empty fibres of $\varphi$ are precisely the cosets of $K$ in $G$. (Again we have to specify non-empty fibre, since $\varphi$ may not be a surjection.)

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Thanks for your detailed answer! – fourdriver01 Aug 25 '12 at 23:26

First and foremost, $\phi$ need not be a bijection. It is not the elements of $G$ that correspond Bijectively to the elements of $\phi(G)$ but the fibres, which as you correctly guess, are the sets of the form $aK$, where $a$ is a general group element, and $K$ the kernel of $\phi$.

Indeed, as you guess, the "other things" in the coset of $a$ is exactly the elements of $aK$, which is really the set $\{ak$ $|k \in K\}$.

Is it because everything in $aK$ is just the same thing?

It is something like that. Because, the fibers of $\phi$ are indeed the sets $aK$ and it is these fibres which correspond Bijectively to $\phi(G)$.

In a little while, Artin will discuss the first Isomorphism theorem, which refines the crude notion of "making all the elements of $aK$ the same thing".

Also, the elements of $G$ are indeed distinct, since $G$ is also a set - as you notice - but it is their images that are not distinct.

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Thanks! I will read ahead! – fourdriver01 Aug 26 '12 at 0:21

I think Brian's answer is excellent and right on the money. In short: whenever we have any function $\,f: A\to B\,$, for any $\,b\in B\,$ we define the fiber over $\,b\,$: $$f^{-1}(b):=\{a\in A\;|\;\;f(a)=b\}$$
The set of all fibers is a partition on the domain $\,A\,$, and it thus determines both an equivalence relation on A -- as any other partition on any set, the equivalence relation in this case being $$\forall\,\,a,b\in A\,\,,\,\,a \sim b\Longleftrightarrow f(a)=f(b)$$ -- , and also a bijection between the image of $\,f\,$ and the set of its fibers.
In case of a group homom. $\,\phi:G\to G'\,$ , we get fibers as above, but this time we can say more, as defining $\,N:=\{x\in G\;\;|\;\;\phi(x)=1\}\,$ , with $\,1=\,$ the unit (in this case, in $\,G'\,$) , we get: $$\forall g'\in G'\,\,,\,\phi^{-1}(g')=xN:=\{xn\;|\;\;n\in N:= \ker\phi\}$$ where $\,x\in G\,$ is any element s.t. $\,\phi(x)= g'\,$
Here, the fibers are the cosets (this is a very important concept in group theory!) $\,xN\,$ , which are, as mentioned above, in $\,1-1\,$ correspondence with the image of the homomorphism $$Im(\phi):=\{g'\in G'\;\;|\;\;\exists\,\,x\in G\,\,s.t.\,\,\phi(x)=g'\}$$ and which also are a partition of $\,G=\,$ the domain group.