Conversation started Jul 12, 2018 at 12:40.
user131753
Jul 12, 2018 12:40
@MatheinBoulomenos Would you mind to clarify this point?
would you mind if I introduce subcategories and essential images while doing so?
user131753
@MatheinBoulomenos No problem.
okay, so a subcategory $\mathbf{A}$ of a category $\mathbf{B}$ is kind of what you would expect: it's a category such that $Ob(\mathbf{A}) \subset Ob(\mathbf{B})$ and for each $A,B \in Ob(\mathbf{A})$, you have $\mathrm{hom}_{\mathbf{A}}(A,B) \subset \mathrm{hom}_{\mathbf{B}}(A,B)$ and the identities in $\mathbf{A}$ are the same as the ones in $\mathbf{B}$ and composition is just restricted from the composition in $\mathbf{B}$
For that for any subclass $X \subset Ob(\mathbf{B})$ we can just define a subcategory $\mathbf{X}$ of $\mathbf{B}$ by saying that $Ob(\mathbf{X}) := X$ and $\mathrm{hom}_{\mathbf{X}}(A,B) := \mathrm{hom}_{\mathbf{B}}(A,B)$ for all $A,B \in X$
and we take the same composition as in $\mathbf{B}$ restricted to the objects in $X$
that's called the full subcategory with objects $X$
note that inclusion of objects and morphisms gives an embedding $\mathbf{A} \to \mathbf{B}$
for any subcategory $\mathbf{A}$ of $\mathbf{B}$
if that embedding is full, then $\mathbf{A}$ is called a full subcategory
that's where the name "the full subcategory with objects $X$" comes from
suppose you have a functor $F:\mathbf{A} \to \mathbf{B}$ and say that $F$ is full for simplicity then we define the essential image of $F$ to be the full subcategory with the following class of objects $X:=\{B \in Ob(\mathbf{B}) \mid \exists A \in Ob(\mathbf{A}), F(A) \cong B$
not that a full and faithful functor $F:\mathbf{A} \to \mathbf{B}$ is isomorphism-dense iff the essential image is just $\mathbf{B}$ (this is just a restatement of the definition)
are you following? @user170039
user131753
Jul 12, 2018 12:56
@MatheinBoulomenos Yeah. Sure.
okay suppose $F$ is fully faithful and let's denote the essential image by $\mathrm{im}_{ess}(F)$, then $F$ factors over the inclusion $\mathrm{im}_{ess}(F) \hookrightarrow \mathbf{C}$
now by definition, the factorized version $F: \mathbf{A} \to \mathrm{im}_{ess}(F)$ is full and faithful and isomorphism-dense, so an equivalence
so if we suppose we know that an equivalence reflects identities iff it is an embedding, we get that $F:\mathbf{A} \to \mathrm{im}_{ess}(F)$ is an embedding iff it reflects isomorphisms
user131753
@MatheinBoulomenos Sorry to interrupt but what is $\mathbf{C}$?
ah lol
that's supposed to be a $\mathbf{B}$, sorry
too late to edit
user131753
@MatheinBoulomenos I see. Then it's fine.
but since the inclusion $\mathrm{im}_{ess}(F) \hookrightarrow \mathbf{B}$ is obviously an embedding and reflects identities, and the original $F$ is the composition $\mathbf{A} \to \mathrm{im}_{ess}(F) \hookrightarrow \mathbf{B}$, we can get the statement that reflecting identities is equivalent to embedding for general full and faithful functors
the other direction is of course clear since equivalences are full and faithful
that's what I meant by saying that the two versions with "equivalences" or just "full and faithful" are basically equivalent
so the idea is just to throw away all objects that prevent the full and faithful functor from being an equivalence
like you can get a surjection from every function by restricting to the image
okay? @user170039
user131753
Jul 12, 2018 13:07
@MatheinBoulomenos How do we get that? If we assume that an equivalence reflects identities iff it is an embedding then (since a fully faithful functor reflect isomorphisms) we can conclude that an embedding reflects isomorphisms. But how do we get the converse?
ah lol
forget that I wrote isomorphisms
corrected version:
"so if we suppose we know that an equivalence reflects identities iff it is an embedding, we get that $F:\mathbf{A} \to \mathrm{im}_{ess}(F)$ is an embedding iff it reflects identities"
user131753
@MatheinBoulomenos Now it's ok.
user131753
(To me at least.)
I agree that you don't use isomorphism-dense anywhere in the proof so it's bit strange that they included it in the exercise statement
user131753
@MatheinBoulomenos Yes. That's what motivated me to write the post. I thought that I must be missing something.
Jul 12, 2018 13:14
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Q: How to solve $ f\left(\sqrt[3]{1 - z^3}\,\right)^2 = 1 - f(z)^2 $?

mickHow to Find analytic $f(z)$ such that $$ f\left(\sqrt[3]{1 - z^3}\,\right)^2 = 1 - f(z)^2 $$ Koenigs function can not be used here So I am stuck. How does the riemann surface look like ?

user131753
@MatheinBoulomenos So if we try to flesh out the details of this comment I think what you meant is something as follows: Let $F:\mathbf{A}\to\mathbf{B}$ be a fully faithful functor then since $F$ can be written as the composition of two embeddings, i.e. $H\circ G$ where $G:\mathbf{A}\to\operatorname{im}_{ess}(F)$ and $H: \operatorname{im}_{ess}(F)\to\mathbf{B}$ and since both of them reflects identities, $F$ also reflects identities.
user131753
@MatheinBoulomenos So the key observation is that any full and faithful functor can be written as the composition of an equivalence and an embedding.
and the other direction works as well
no wait
that's not true
any full and faithful functor is the composition of an equivalence and an embedding
$\mathbf{A} \to \mathrm{im}_{ess}(F)$ need not be an embedding
user131753
@MatheinBoulomenos Yeap. Noted that.
Jul 12, 2018 13:24
I think that's a nice statement to have in general, so it's good you pointed that out
user131753
@MatheinBoulomenos In my comment here also the edit should be done accordingly.
Note that $\mathrm{im}_{ess}(F) \to \mathbf{B}$ reflects identities by construction
so we can use this decomposition to deduce the statement from the exercise for general full and faithful functors from the (a priori weaker) statement just for equivalences, that's why the formulations are "basically equivalent" in some vague sense
user131753
@MatheinBoulomenos However if $F$ is a full and faithful functor that reflects identities then can we show that $F$ is an equivalence?
user131753
Only assuming that if $F$ is an equivalence then $F$ is an embedding iff it reflects identities.
user131753
@MatheinBoulomenos I don't think they are logically equivalent.
 
Conversation ended Jul 12, 2018 at 13:39.