Brainlet here, is this an example of a monomorphism?

Say A = {1,2} and B = {6,7,8,9}, and let f : A -> B be defined by f(1) = 7 and f(2) = 9. It shouldn't be hard to see that f is injective.

Pick any other set C and a function g : C -> A. Say we take C = {x,y,z} and g(x) = 1, g(y) = 1, and g(z) = 2. Then the composition fg is the map which gives fg(x) = 7, fg(y) = 7 and fg(z) = 9.

Whenever you draw this out into a picture you'll quickly see,

g(x) = 1
g(y)=1

and x=/=y. so in my mind this is not 1-to-1 and since monic morphism is suppose to be a generalization of injection, I would think this alone fails the condition of a monomorphism.

Can someone explain this to me?

What is my misunderstanding?

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you never defined g_1 and g_2 brainlet, only a single function g

and yes f is a monomorphism because if
f g_1 = f g_2 for two functions g_1, g_2 from C to A then
f g_1(c)=f g_2(c) for all c in C
since f is injective this implies g_1(c)=g_2(c) for all c in C

Reading. Will reply once my brainlet mind thinks this through

I'm still a little confused. How is f a monomorphism?

g maps x to 1
g maps y to 1

this means g maps two different elements to the same element.,

I thought by the definition of injecticvity (in set theory) if f(a)= f(b) => a = b, and here g(x) = g(y) but x and y are two different elements so its no injective and thus since monomorphism is a generalization of injection it shouldn't be a monomorphism either.

Where is my understanding (besides being a brainlet)?

Is monomorphism only concerned with what the arrows are doing, not what the objects its mapping the objects to?

you're still not even using the definition of monomorphism

you need to show that f g_1 = f g_2 implies g_1 = g_2

all you've defined is a single g, and it doesn't matter if that map is injective or not

f g_1 = f g_2 implies f g_1 (x)= f g_2 (x) for all x in C

so f(g_1(x))=f(g_2(x)) for all x in C

if f is injective then this implies g_1(x)=g_2(x) for all x in C, and so g_1= g_2, and so f is a monomorphism

So if I am not using the definition of monomorphism then how is f a monomorphism in my example?

Okay, let me define it in this new way:

I will call this "Example 1"

it's the same as before except now g,g_1, g_2: C->A and f: A->B.

A={1,2}
B={6,7,8,9}
C={x,y,x}

g_1(x) = 1
g_2(x) = 1

g_1(z) = 2
g_2(z)=2

g(x)=1
g(y)=1


f is the same,

f(1)=7
f(2)=9

Now is this monomorphic?

Or are all the elements in the category
{x,y,z,12,6,7,8,9} and g_1 and g_2 needs to map them too?

Now let's look at example 2:

This is NOT monomorphic right?

g_1(x) = 1
g_2(x) = 1

g_1(z) = 2
g_2(z)=9

g(x)=1
g(y)=1

Lastly,

So it is possible for f to not be injective but for f to be a monomorphism:?

FUCK. I meant to say is g injective? clearly f is.

So is g injective / a monomorphism?

because g doesn't seem injective in example 1 since it takes two different elements and maps them to the same element.

So is g monomorphic?

g is not injective since you wrote g(x)=g(y)=1

in the category of sets monomorphisms are exactly the injective functions

OOOKKK so g is neither injective nor a monomorphism......

but fuck how is it not a monomorphjism now?

would this be a monomorphism

g_1(x) = 1
g_2(x) = 1

?

that is a morphism that maps two different elements to the same element.


I'm fucking confused because
it seems to satisfy the definition of mono but fail the condition of injection.


because in example 1: fog1 = fog2 yet g1 =/= g2

I mean to write *****

g_1(x) = 1
g_1(y) = 1

is this a monomorphism?

at this point i have absolutely no idea what function you're even asking is a monomorphism or not but

if
g(x)=g(y)=1
and you pick two functions g_1, g_2 from {a,b} to {x,y} with
g_1(a)=g_2(b)=x
and g_1(b)=g_1(a)=y

then g g_1 = g g_2 but you don't have g_1= g_2

Okay in your new example this would fail the definition of injection AND monomorphism?

yes, g is neither an injection (because g(x)=g(y)=1) nor a monomorphism (since gg_1= gg_2 but g_1 is not g_2)

okay.... so going back to my original problem....


Say A = {1,2} and B = {6,7,8,9}, and let f : A -> B be defined by f(1) = 7 and f(2) = 9. It shouldn't be hard to see that f is injective.

Pick any other set C and a function g : C -> A. Say we take C = {x,y,z} and g(x) = 1, g(y) = 1, and g(z) = 2. Then the composition fg is the map which gives fg(x) = 7, fg(y) = 7 and fg(z) = 9.

Whenever you draw this out into a picture you'll quickly see,

g(x) = 1
g(y)=1


clearly g is not injective (it fails the definition of injection), so how does this fail the definition of monomorphism now?

is it because g is one map and monomorphism requires 2 maps?


what confuses me about the definition of monomorphism is it basically

says f o g1 = f o g2 => g1 = g2

and in my example you see

f(g(x)) = 7
f(g(y)) = 7

and this says since f(g(x)) = f(g(y)) => x = y and thus it's monomorphic but thats not true since x=/=y so I am trying to understand where the definition shows this isn't a monomorphism

in this post, f IS a monomorphism and g ISNT

>clearly g is not injective (it fails the definition of injection), so how does this fail the definition of monomorphism now?
because you can define g_1 and g_2 like i did in the previous post

two functions g_1, g_2 from {a,b} to {x,y} with
g_1(a)=g_2(b)=x
and g_1(b)=g_1(a)=y

then g g_1 = g g_2 but you don't have g_1= g_2

>and this says since f(g(x)) = f(g(y)) => x = y
no it doesn't, f being injective gives g(x)=g(y), not x=y

Okay, this is clearing up a ton of misconceptions.

Lastly, when you say you don't have g_1 = g_2 what do you mean by this?

Does this mean g_1 and g_2 are mapping two different elements to the same element and since the two different elements being mapped aren't equal to each other, then g_1 and g_2 aren't the same?

>Lastly, when you say you don't have g_1 = g_2 what do you mean by this?
when i say we dont have g_1 = g_2 thats because
g_1(a)=x and g_2(a)=y , so g_1 and g_2 don't agree on a so they're not the same function

>Does this mean g_1 and g_2 are mapping two different elements to the same element and since the two different elements being mapped aren't equal to each other, then g_1 and g_2 aren't the same?
no, not because they send different elements to the same elements, its because they send the same element to two different elements

please either go read a proof that monomorphisms in Set are exactly injections or try to write one down yourself, you really shouldn't go any further if you can't grapple with that yet

Ok thanks for your responses. I am going to take a break, read over what you wrote and see if I understand it. I think what you wrote set me in the right direction.

Thanks so much for your help.

I think I finally understand your example.

g(x)=1
g(y)=1

g_1(a) = x
g_2(b) = x

g_1(b)=y
g_2(a)=y

g g_1 = g g_2 but g_1 =/= g_2

so I can conclude g is not monomorphic.

good work

Yes, monomorphism is only concerned with arrows. In fact, you really should make clear in which category you're working when you mention monomorphisms. For instance, there are non-injective monomorphisms in the category of rings.

Thanks.

For epimorphism: for all morphisms g1, g2 : Y → Z,

[math]{\displaystyle g_{1}\circ f=g_{2}\circ f\Rightarrow g_{1}=g_{2}.} g_{1}\circ f=g_{2}\circ f\Rightarrow g_{1}=g_{2}.[/math].

It sounds like the reverse of monomorphism. Can you show me an example of an epimorphism using the category of sets and how does this definition (applied to sets) guarantee that every element in the range will be mapped from some element in the codomain?

I said that weird, how does the definition of epic guarantee that g is surjective for the category of sets? that isn't immediately obvious to me from the definition, also can you provide an example of epic using the category of sets?

^ That is what I meant to write.

> how does the definition of epic guarantee that g is surjective for the category of sets?
en.wikipedia.org/wiki/Epimorphism#Examples