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Apr
24
comment Subcategory of category of Module satisfies SSA?
@henry answer 2 applies to every category of modules. About answer 1 it seems that the same argument could be applied to the case of $\mathbb Q[x]$-modules (direct product of infinite free modules, copies of $\mathbb Q[x]$, need not to be free). Note that $\mathbb Q[x]$ is an infinite dimensional $\mathbb Q$-algebra.
Apr
23
comment Subcategory of category of Module satisfies SSA?
What is exactly the question in 2? Are you asking if the inclusion functor verifies the solution set condition?
Apr
18
comment Relation between Cartesian closed category and Lambda Calculus
Of course my previous comment applies if by lambda-calculus you mean the untyped lambda-calculus. Did you meant symply typed lambda-calculus instead?
Apr
18
comment Relation between Cartesian closed category and Lambda Calculus
I think that question 2) is ill posed: lambda calculus cannot express the structure of a cartesian closed category. On the other hand you can interpret lambda calculus in a cartesian closed category.
Apr
16
comment How to prove this Category Theory theorem?
@user158083 Give me some time, I'll write another answer where I'll address this question.
Apr
15
comment How to prove this Category Theory theorem?
Feel free to ask if you need additional clarification or details.
Apr
15
comment How to prove this Category Theory theorem?
@user158083 I've made some changes hopefully they'll get you a better understanding of where the problem lies. Meanwhile I've also add a counter-example to one of the wrong implications, finding a counter-example to the other implication it's a little harder since it would require to find a functor $H$ which has a left adjoint and a functor $F$ whose object part agree with such left adjoint, but the arrow part does not.
Apr
15
comment How to prove this Category Theory theorem?
If you'll find stuck in finding the counter-example feel free to ask. I'll try to post them as soon as possible, but I believe that it would more instructive if you tried to find them by yourself first.
Apr
9
comment Free monoidal category over a set
@user329838 are you referring to the first line in the answer? If that's the case they are similar because in strict categories you don't simply require that the $n$-fold product of objects to be equal, you also require that associators and left and right unit are identities. In your question you didn't add such a requirement so the free-strict-category is not exactly the category you describe either.
Apr
8
comment Free monoidal category over a set
Just a speculation.
Apr
8
comment Free monoidal category over a set
If you work with $1$-categories I guess so. Nonetheless if you instead want to consider the $2$-category of monoidal categories and monoidal functor between them.... well I think the story could be different: I think (if not entirely sure) that in that case the two categories should be equivalent, but then you shouldn't have a unique functor $\bar f$, only one unique up to natural isomorphism.
Apr
8
comment Free monoidal category over a set
Not if you are working in the $1$-category of monoidal categories and functor between them. To understand why try to answer this question: if your category were initial it should exist a functor into my category, what should be the image of an object $x_1 x_2 x_3$ through this functor? Should it be $(x_1 x_2)x_3$ or $x_1 (x_2 x_3)$?
Feb
24
comment Which is the difference between the two notations?
Perfect answer, allow me also to rephrase some of the thing you have written above (sometimes rephrasing helps understand better). We have that $g \in N_G(H)$ if and only if the restriction of $\phi_g$ to $H$ give an endomorphism, actually an automorphism, of $H$ (i.e. an ${\phi_g}_{|H} \colon H \to H$). Instead $g \in C_G(H)$ if and only if not only ${\phi_g}_{|H}$ is an $H$-automorphism but in particular ${\phi_g}_{|H}=\text{id}_H$. This is just a different way to say what anomaly already said.
Feb
23
comment Tensor product of enriched functors
Could you cite the reference where you've seen this tensor product between functors?
Feb
23
comment Equivalence Relations on Products
@mil10 who said that $X$ is a subgroup? By the way for the definition of $n$-tuples take a look to wikipedia's page which can give more details than the one I could give in a comment.
Feb
19
comment Equivalence Relations on Products
$X$ is not at all a subset of $G$, it is a subset of $G^n$: its elements are $n$-tuples of elements of $G$ whose product is equal to $1$.
Feb
17
comment Defining Surjection as an Implication
@RobArthan cool I didn't notice that, thanks.
Feb
15
comment Does one need to learn set theory before learning category theory?
You're welcome! ;)
Feb
15
comment Does one need to learn set theory before learning category theory?
To that you could add the following piece of text from the hisorical notes in the first chapter of MacLane's Categories for the working mathematicians "Categories, functors, and natural transformations themselves were discovered by Eilenberg-Mac Lane in their study of limits (via natural transformations) for universal coefficient theorems in Cech cohomology."
Feb
15
comment Does one need to learn set theory before learning category theory?
@AsafKaragila Well I don't know about that but if look almost everywhere you'll find out that the most ancient paper on categories is the famous "General theory of natural equivalences" where the authors (MacLane and Eilenberg) provide a list of examples from the real of algebra and topology, no logic (with the exception of some remarks on foundations).