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Julius (Aug 06 2023 at 13:22):Exercise 1.1.i.
Consider a morphism f : x → y. Show that if there exists a pair of morphisms g,h: y ⇒ x so that gf = 1x and fh = 1y, then g = h and f is an isomorphism.
Definition 1.1.9. An isomorphism in a category is a morphism f : X → Y for which there exists a morphism g: Y → X so that gf = 1X and fg = 1Y.
To be an isomorphism, there has to exist a g: Y -> X which composes to form the identity arrow id, both before, and after, the arrow f: X -> Y. Is there an example of an arrow which only produces the identity arrow for one of those directions, like f;g = id_x, but not g;f = id_y ?
Evidently, I am supposed to infer that since there is an inverse arrow for both sides, they must be the same arrow. Maybe I can show this through algebraic manipulation, but I wonder if there is some way to know it inherently and immediately, in some way?
f;g = id_x
h;f = id_y
f;g;f = f
h;f;g = id_y;g
h;id_x=id_y;g
h = g
I believe that’s the trick. A simple algebraic manipulation. I’ll just have to think over if there’s an intuitive way to say what it’s saying. I think it’s something to do with the axioms of a category would not permit a concept of left-only or right-only inverse arrow; perhaps due to the associativity of the composition.
Julius (Aug 06 2023 at 13:31):(ii) Show that a morphism can have at most one inverse isomorphism.
Take some morphism f: x -> y.
Assume it has an inverse arrow, g: y -> x, such that f;g = id_x.
Assume there is a second inverse arrow, g’: y -> x s.t. f;g’ = id_x.
An isomorphism is necessarily two-sided.
So
f;g = f;g’
g;f;g = g;f;g’
id_y;g = id_y;g’
g=g’
I mainly need to think over why an inverse is necessarily two-sided, or if it isn’t, it’s just that an isomorphism necessarily is.
John Baez (Aug 06 2023 at 13:31):Evidently, I am supposed to infer that since there is an inverse arrow for both sides, they must be the same arrow. Maybe I can show this through algebraic manipulation...
Yes, there's a famous little trick for doing this, which people usually learn in group theory, and it looks like you found it.
I think of this and similar arguments as "fastest gun in the west" arguments. If two guys say they're the fastest shooter in town, you settle it with a duel.
Suppose g says he's the right inverse of f but h says he's the left inverse of f. Then you let them fight it out! You look at
h f g
(I'll leave out the the semicolons since they're distracting here.)
First h makes his move:
h f g = g
Then g makes his move:
h f g = h
So g and h had to be the same!
When two guys claim they're the fastest gun in the west and they're both right, they're probably the same guy.
John Baez (Aug 06 2023 at 13:33):I mainly need to think over why an inverse is necessarily two-sided, or if it isn’t, it’s just that an isomorphism necessarily is.
There's no reason an inverse is necessarily two-sided. Find two functions f and g such that f is a left inverse of g but not a right inverse. Then g will be a right inverse of f but not a left inverse.
It's just the definition of isomorphism that says it's a morphism with a two-sided inverse.
John Baez (Aug 06 2023 at 13:33):All this stuff is very fun to think about.
Julius (Aug 06 2023 at 14:35):Thank you very much.
“””
Exercise 1.1.iii. For any category C and any object c ∈ C, show that:
(i) There is a category c/C whose objects are morphisms f : c → x with domain c and in which a morphism from f : c → x to g: c → y is a map h: x → y between the codomains so that the triangle commutes, ie g = hf.
“””
I am slightly confused about “show that there is a category that”, because we are guaranteed to be able to “select” the objects of the slice category - all arrows in C of the form f: c -> x, for any x such that there is an arrow from c to x.
So that part doesn’t require us to prove anything. Right?
As for the arrows between the objects, I got the impression we are not selecting arrows from the original category C, because there’s no guarantee to be an arrow h: x -> y, for f: c -> x, g: c -> y. It just depends on what arrows the category C has.
So we do not select pre-existing arrows, we actually just declare the existence of them. For any objects f: c -> x and g: c -> y in the slice category C under c, we state there is an arrow h: x -> y, so that g = f;h.
Is that correct?
It seems like there’s nothing to “show”. We can always construct a slice category. Or should I prove that all the axioms of a category (identities, composition, associativity) still hold, under this construction?
Morgan Rogers (he/him) (Aug 06 2023 at 14:40):No, the arrows are taken from the original category. Just like a group homomorphism is a function between groups satisfying certain conditions, a morphism in the slice category is a morphism from the original category satisfying a particular condition.
The exercise is indeed to check the axioms of a category: they have told you what the objects and morphisms should be, but what are the identities? What does composition look like? Is it still associative? After a certain point these checks will become "obvious" enough that you won't need to check them by hand, but exercises like this should help you internalise the definition of a category.
John Baez (Aug 06 2023 at 16:29):Note that this exercise is going beyond the basic sort of proof where the book tells you a proposition and you prove it's true. Here the work involves guessing the proposition and then proving it. For example: the book is not telling you how composition of morphisms in the slice category works, and until you say what that is, it makes no sense to prove it's associative.
In the language of modern math, an elementary sort of textbook problem is asking you to verify a property. Here you are being asked to choose a structure, state the relevant property of that structure, and then prove it.
John Baez (Aug 06 2023 at 16:32):Later still you may be asked to choose some 'stuff', then choose some structure on that, and then verify some properties.
Jencel Panic (Aug 08 2023 at 10:11):@John Baez -- best math explanation in the West :smile:
John Baez (Aug 08 2023 at 10:57):Thanks so much! I figured this might be what was bothering you.