I’m confused how we are using the forgetful functor to derive the product in the category of vector spaces. How do we know if P is a product of V and W in the category of vector spaces that F(P) is a Cartesian product of V and W in the category of sets where F is the forgetful functor?
This seems to be what the argument relied on, and furthermore we know that products are not unique (may be unique up to isomorphism but they are not literally unique) so if P is a product of vector spaces that is not literally a set theoretic Cartesian product of V and W then when we apply the forgetful functor F we won’t get a Cartesian product and I’m not sure how the argument then works.
I’m still not sure what the forgetful functor tells us though. If P is a product in the category of vector spaces, how do we know that F(P) is a product in the category of sets? I understand that this is true but not why the forgetful functor implies it.
We have the commutative diagram that defines the product in the category of vector spaces. Apply the forgetful functor: now, we have a new commutative diagram, but in the category of sets. ...But this is the commutative diagram of the product in the category of sets! This tells us what we should do to get the product in the category of vector spaces: first, produce the product in the category of sets (e.g., the Cartesian product will do); second, find a way to lift it to a vector space.
Right I see how this tells us what we would expect the product in the category of vector spaces(Vec) to look like from what you wrote, but I guess my problem is that this doesn’t seem like a proof of that. After applying the forgetful functor to the commutative diagram for the categorical product in the Vec you get a commutative diagram looking like a product diagram in the category of sets(Set); however, only morphisms that were linear transformations in Vec had the forgetful functor applied to them(as those are the morphisms in that category). Therefore, F(P) may look like a product in the Set, but we need it to satisfy a commutative diagram saying something about all functions(morphisms in Set) out of some set into some sets, not just functions that happen to be the image of linear transformations after taking the forgetful functor. Of course this suggests what the product P in Vec looks like, but doesn’t by itself prove it must be (isomorphic to) that.
That’s my confusion, I might just be missing something sorry. Thank you for the help!
You are, of course, right that the commutative diagram is missing a bunch of objects and morphisms. But we know that there is something that will satisfy the truncated diagram *and more*. Now, of course, you could worry that maybe there is something some *other* object that is not equinumerous with this one that satisfies the truncated diagram. Here, admittedly, we need a bit more than just the existence of the forgetful functor---we need to know something about the category of sets. Namely, that most of those objects and morphisms are actually superfluous.
I liked this post a lot! One minor point though on initial objects, https://youtu.be/5uI0uJpsEhI shows the initial object for vector spaces as being R rather than a single point as you described.
The initial object for R-Vect can't be R. An initial object is an object I where Hom(I, A) is a single point for any A in the category. But Hom(R, V) = V, not a single point.
Lukas is absolutely right, but here's an easier way to see it: there are infinitely many linear transformations between any two 1-dimensional spaces (corresponding to multiplication by each possible scalar)---ergo, they can't be either initial or terminal objects!
Thanks, I must have misunderstood the point in the video at 11min in. From that it seemed that using the zero vector as initial object results in all vectors being the zero vector since the morphisms in Vect_R are linear maps, while using R as the initial object gives you all vectors. The map used to define a given vector is f: R -> V with f(1) = \vect{v}.
If you use the vector space containing only the zero vector then, indeed, your only option for a linear map into some other space V is that it sends the zero vector to the zero vector in V. But that's exactly what you want! The whole point of an initial object is that it should allow only *one* morphism.
I’m confused how we are using the forgetful functor to derive the product in the category of vector spaces. How do we know if P is a product of V and W in the category of vector spaces that F(P) is a Cartesian product of V and W in the category of sets where F is the forgetful functor?
This seems to be what the argument relied on, and furthermore we know that products are not unique (may be unique up to isomorphism but they are not literally unique) so if P is a product of vector spaces that is not literally a set theoretic Cartesian product of V and W then when we apply the forgetful functor F we won’t get a Cartesian product and I’m not sure how the argument then works.
Loved the post, just something I was confused by.
It need not be exactly the Cartesian product---it is just the Cartesian product up to isomorphism. That's enough!
Gotcha, thanks!
I’m still not sure what the forgetful functor tells us though. If P is a product in the category of vector spaces, how do we know that F(P) is a product in the category of sets? I understand that this is true but not why the forgetful functor implies it.
We have the commutative diagram that defines the product in the category of vector spaces. Apply the forgetful functor: now, we have a new commutative diagram, but in the category of sets. ...But this is the commutative diagram of the product in the category of sets! This tells us what we should do to get the product in the category of vector spaces: first, produce the product in the category of sets (e.g., the Cartesian product will do); second, find a way to lift it to a vector space.
Right I see how this tells us what we would expect the product in the category of vector spaces(Vec) to look like from what you wrote, but I guess my problem is that this doesn’t seem like a proof of that. After applying the forgetful functor to the commutative diagram for the categorical product in the Vec you get a commutative diagram looking like a product diagram in the category of sets(Set); however, only morphisms that were linear transformations in Vec had the forgetful functor applied to them(as those are the morphisms in that category). Therefore, F(P) may look like a product in the Set, but we need it to satisfy a commutative diagram saying something about all functions(morphisms in Set) out of some set into some sets, not just functions that happen to be the image of linear transformations after taking the forgetful functor. Of course this suggests what the product P in Vec looks like, but doesn’t by itself prove it must be (isomorphic to) that.
That’s my confusion, I might just be missing something sorry. Thank you for the help!
You are, of course, right that the commutative diagram is missing a bunch of objects and morphisms. But we know that there is something that will satisfy the truncated diagram *and more*. Now, of course, you could worry that maybe there is something some *other* object that is not equinumerous with this one that satisfies the truncated diagram. Here, admittedly, we need a bit more than just the existence of the forgetful functor---we need to know something about the category of sets. Namely, that most of those objects and morphisms are actually superfluous.
I’m not sure what makes them superfluous, but it seems I’m missing some of the additional information. Thanks again for the thoughtful responses :)
Small correction: you wrote ϕ∘f^-1: B—>D but I believe you meant Φ∘f^-1: B—>D instead.
ϕ∘f^-1 is actually a morphism from B to A.
Yes: I accidentally misplaced a \varphi, rather than \phi. Thank you!
I liked this post a lot! One minor point though on initial objects, https://youtu.be/5uI0uJpsEhI shows the initial object for vector spaces as being R rather than a single point as you described.
Apologies, from the replies it looks like I was misunderstanding here and probably mixing up initial objects and generators.
The initial object for R-Vect can't be R. An initial object is an object I where Hom(I, A) is a single point for any A in the category. But Hom(R, V) = V, not a single point.
Lukas is absolutely right, but here's an easier way to see it: there are infinitely many linear transformations between any two 1-dimensional spaces (corresponding to multiplication by each possible scalar)---ergo, they can't be either initial or terminal objects!
Thanks, I must have misunderstood the point in the video at 11min in. From that it seemed that using the zero vector as initial object results in all vectors being the zero vector since the morphisms in Vect_R are linear maps, while using R as the initial object gives you all vectors. The map used to define a given vector is f: R -> V with f(1) = \vect{v}.
If you use the vector space containing only the zero vector then, indeed, your only option for a linear map into some other space V is that it sends the zero vector to the zero vector in V. But that's exactly what you want! The whole point of an initial object is that it should allow only *one* morphism.