Consider the canonical
projection π:Q→Q/Z3. We aim to show that this is a monomorphism despite it not being
injective
For anyone unfamiliar,
Q/Z can be roughly though of as
Q∩[0,1).
(Note that
π is
surjective and therefore epic, meaning that it also gives an example of a
morphism which is both monic and epic but not an
isomorphism)
To show
π is monic, first start by taking
morphism f,g:A→Q for some divisible
abelian group A. We assume that
πf=πg and must show that
f=g.
First we show that
f and
g always differ by some integral amount. Take
a∈A. We know that
π(f(a))=π(f(b)). Since
π acts by collapsing
equivalence classes x+Z to a single value, then
π(f(a))=π(f(b)) entails that
f(a) differs from
f(b) by some integral amount; that is,
f(a)−g(a)=kfor some k∈Z
Using this, we can show that
f and
g agree on nonzero values. Take
z∈A other than
zero. By the above know
f(z)−g(z)=m
for some integral
m. Now assume towards contradiction that
f(z)=g(z); ie, that
m=0. Since
A is divisible we can ‘divide’
a by
2m to produce a value
b so that
b⋅2m=a. Then since
f,g are
homomorphisms we have
f(b)−g(b)=2mf(a)−g(a)
we know already that
f(a)−g(a)=m, so the whole fraction is
m/2m=1/2. Hence
f(b) and
g(b) differ by a non-integral value. But we’ve already established that this is impossible, so we must reject our assumption that
f(z)=g(z).
This holds for all nonzero
z, and hence
f and
g must agree except perhaps at
zero. And we can show that they agree at
zero by choosing any nonzero
y∈A; then we get:
f(0)=f(y−y)=f(y)−f(y)=g(y)−g(y)=g(y−y)=g(0)
So
f and
g agree everywhere, so
f=g.
The
BCC instructor explains the situation by saying that although
π is not
injective, it is “
injective on the important part of
Q" (ie, on the integral part of numbers) from which the rest of
Q is induced.
I also offer the following illustration. (This illustration is intended primarily for aid with conceptual understanding and isn’t presented with the claim of high rigor)
Here we have
Q pictured as
ω-many copies of
Q/Z which are “stacked along the y axis”. The
projection π conflates these copies or “stacks them along the z axis” (ie, orthogonal to your computer screen).
On the left we have some element
a∈A which is being sent (towards contradiction) by
f,g to different elements
f(a),f(g) of
Q.
The fact that the post-compositions
πf and
πg are equal means that after “flattening” by
π we know that
f(a) and
f(g) must be sent to the same value in
Q/Z; hence the images
f(a) and
g(a) must have differed by some integral amount—some integral number of “copies of
Q/Z". Call this number
m. Since
A is divisible we can “divide”
a by
2m to produce
b with
b⋅2m=a. And since
f,g are
homomorphisms this will be reflected: the images
f(b) and
g(b) will similarly be “divided” by
2m, so that their difference is now only
1/2. This places them within the same copy of
Q/Z, meaning that under
π they would be sent to different values. Contradiction.
Since by assumption , then 2. Applying monicity of we get then that and thus that . Hence is injective!
Assume that ; we must show that .
First let , then:
This holds for all , so we have that and agree over . We need to extend this to all of .
So take some , and write it as for . Note then that
due to being a homomorphism; the same holds for . Then we can proceed by the following chain of equalities:
Hence . Since this holds for all , then .
Abstractly what happened is this. We wanted to lift the agreement of and on up to all of . We did show by using the structure of rings to prove that since and are homomorphisms then their behaviour on all of is decided by their behaviour on (equation (A)). With a little algebra this became .
The BCC instructor explained the situation like this: although is not surjective, it is “surjective onto the important part of " (ie, onto ); from this ‘important part’ the rest of is decided. It is this fact which makes monic.
Here we have pictured as -many copies of which are “stacked along the y axis”. The projection conflates these copies or “stacks them along the z axis” (ie, orthogonal to your computer screen).
On the left we have some element which is being sent (towards contradiction) by to different elements of .
The fact that the post-compositions and are equal means that after “flattening” by we know that and must be sent to the same value in ; hence the images and must have differed by some integral amount—some integral number of “copies of ". Call this number . Since is divisible we can “divide” by to produce with . And since are homomorphisms this will be reflected: the images and will similarly be “divided” by , so that their difference is now only . This places them within the same copy of , meaning that under they would be sent to different values. Contradiction.
This category consists of two identity morphisms and one non-identity . The non-identity is both monic and epic (why?), but is not iso since it has no inverse.