If you remember the derivation, the formula for the derivative of an implicit function only works if the original function has non-zero $y$-derivative at that point. It doesn't make sense otherwise.

A measure space is a space with a measure?

@nbro sure

It is a measurable space with a measure, to be precise

@Thorgott Yes I remember, but a derivative is a limit, and if the part above and part below are equal, than there would be no problem

What is the difference between a measurable space and a measure space?

A probability space is a measurable space with a measure called the probability measure, which is often denoted as $\mathbb{P}$, and is often called the probability distribution.

I mean, the assumption of the theorem is stated to prevent the occurence of a denominator=0, but in the example above, this wouldn't be occur after we consider also the numerator

Yes, but you are talking about the explicit formula that expresses the derivative of the implicit function in terms of the derivatives of the original function, no? @Shootforthemoon

yes, I know

But cannot we keep the derivatives as limits?

we would have 0+/0+

@nbro A measurable space is a tuple $(X,\mathcal{M})$, where $X$ is a set and $\mathcal{M}$ is a $\sigma$-algebra on $X$. A measure space is a triple $(X,\mathcal{M},\mu)$, such that $(X,\mathcal{M})$ is a measurable space and $\mu$ is a measure on $(X,\mathcal{M})$.

@Shootforthemoon I don't think I understand what you mean

So, a measurable space does not possess a measure? Weird terminology

Why not calling it $\sigma$-algebra space?

So, a probability space is a measure space, because it possesses a measure

$\sigma$-algebra space does not sound like a handy term. Measurable space is a perfectly fine terminology, because the $\sigma$-algebra dictates which sets can be measured (i.e., are measurable). A measure then specifies how we measure them.

@Thorgott Yesterday, when you said that a probability distribution is a measure over a measurable space such that the measure is 1, you meant that $P(\Omega) = 1$, where $\Omega$ is the sample space?

I know that a probability measure must assign 1 to the event that contains all outcomes (i.e. the sample space)

@MikeMiller This isn't a real issue; choose a lift of the sphere upstairs which bounds a ball containing no other lift. Then the covering space restricted to the ball is a covering space onto image, so the image of the ball downstairs has to be a ball

This is just saying if $M \to N$ is a covering map of $3$-manifolds, $M$ is irreducible then $N$ is also irreducible

At least, I think it isn't a real issue, and hopefully I'm not being dumb

@Thorgott In your definition of a measurable space, in the case of a probability space, $X$ is the sample space and $M$ is the event space?

I wrote this, not sure it may work

at the end x and y go to zero at the same speed

An event space is a $\sigma$-algebra.

That's right, @nbro.

@Shootforthemoon You wrote $\partial y$, but it looks like you're calculating the derivative wrt $x$. Either way, that limit at the end does not exist.

sry, I also forgot a minus at the end

Yesterday, @Thorgott defined a probability distribution as a synonym for probability measure. In the case of the Kullback-Leibler divergence, does this latter calculate the divergence between probability measures, or what? What does it mean to calculate the divergence between probability measures, anyway? The KL is usually defined as the divergence between "probability distributions", but maybe they refer to CDFs (or something else), rather than probability measures?

@Thorgott no, sure, I didn't write well the formulae,

Statisticians are loose with the word "probability distribution". They usually mean a random variable, it's cdf/pdf, or sometimes worse: a sample off the random variable, when they say that.

But this is not inconsistent with Thorgott's statement because a random variable defines a probability measure. You don't necessarily need to know that.

@BalarkaSen Well, he said that not all random variables have an associated distribution, if I recall correctly

I said that not all random variables have a pmf or pdf.

@BalarkaSen Yes, this is exactly what I thought!

I did say every random variable induces a distribution.

Ok

So, does every r.v. also induce a CDF?

What do you think?

I have no idea

Probably yes

But this is just a guess

Well you know what those words mean, right? Then it's a 30 seconds check to go through the definitions in your head and tell us if it's true or not

When someone talks about CDFs, I think about a function that assigns probabilities to certain parts (subsets) of the graph

Unless you don't know what some of those words mean, in which case: which word out of "r.v." and "cdf" are you confused about?

OK. Suppose I give you the uniform distribution on $(0, 1)$, a statistician's favorite distribution. What's this guy's cdf?

Even simpler, uniform distribution on $\{0, 1\}$, if you like!

The cdf would assign a probability to subsets of (0, 1). At least, this is the first thing that comes to my mind. For example, the cdf would tell us what's the probability that P(X < 0.5), whatever X is

It almost seems that the cdf is a synonym for probability measure

Getting close. Cdf is a "function" right, by virtue of it's name (cumulative distribution function)? So what is your function?

P(X < 0.5) is a good example. X is Unif(0, 1), not anything arbitrary.

Also, for the record, the Kullback-Leibler divergence is indeed defined between distributions (that fulfill certain properties).

@BalarkaSen A function from subsets of (0, 1) to [0, 1]

@Thorgott Now I've been more careful

@nbro That's not right.

@Thorgott So, the KL divergence measures the divergence between probability measures?

@nbro you are actually thinking about the distribution itself, not its cdf

What's the difference between cdf and probability measure, then?

You cannot graph a function from subsets of (0, 1) to [0, 1]. But people draw pictures of CDFs

The CDF of a random variable $X$ is a function $F_X : \Bbb R \to [0, 1]$, $F_X(x) = \Bbb P(X \leq x)$. That's definition.

@Thorgott Where am I thinking about distribution and not cdf?

@BalarkaSen Why is the domain the real numbers?

Because a random variable, classically speaking, takes values in real numbers.

$(X \leq x)$ always makes sense as an event

But the cdf is a function from subsets not from numbers

No, that's the distribution

No. That's where I am telling you that you're wrong.

Ha, sorry, I read "That's right", I forgot the "not"

You're thinking of the measure induced from the random variable (I want to stop calling it distribution for the sake of not confusing terminologies)

So, every random variable has an associated cdf

Yes, correct.

@BalarkaSen You want to stop calling it a distribution because, in statistics, it can refer to pdfs, cdfs or pmfs?

And, even better, the cdf uniquely determines the measure

(I'll follow the convention)

@Thorgott Can you be a little bit more explicit, even though this makes intuitive sense?

The CDF determines the measure induced from the random variable, and the way I'd unpack is that CDF determines the random variable upto null probability sets.

@Shootforthemoon Is this correct? @Thorgott

Can we say that the r.v. also induces the cdf or you just say that the r.v. induces the prob. measure, and why?

So if $X, Y$ are two random variables and $\Bbb P(X \leq x) = \Bbb P(Y \leq x)$ for every $x \in \Bbb R$, "$X = Y$ almost everywhere", which means for every almost every event $E$, $X(E) = Y(E)$, that is to say, the set of events $E$ for which $X(E) \neq Y(E)$ forms a zero probability set.

@nbro I think it's better to not try to understand the measure theoretic counterpart of the whole story at this point. Try picking up the basic probability first.

If you know the cdf, let's call it $F_X$, then you know, by definition $\mathbb{P}(X\le x)$ for all real $x$. Then you also know $\mathbb{P}(x<X\le y)=F_X(y)-F_X(x)$. So, you know the probability that $X$ lands in any half-open interval. All measurable sets are "made up of" half-open intervals, so you know the distribution.

@BalarkaSen I understood all single parts, but I don't get the overall meaning and the "forms a zero prob. set"

@Thorgott Ok, I like this more, thanks

@Shootforthemoon you can not divide limits if they don't exist

what about the limit for x approaching 0 of sinx/x?

So, why do you say that a r.v. induces prob. measure but not a cdf, even though cdf completely determines the prob. measure?

Because "inducing a measure" is a technical well-defined terminology.

That limit exists. I'm saying it wouldn't make sense to write $\lim_{x\rightarrow0}\frac{\sin x}{x}=\frac{\lim_{x\rightarrow0}\sin x}{\lim_{x\rightarrow0}x}$, because that would be $0/0$.

I had actually asked a question regarding the expression "induced by" on this website several months ago

@Thorgott Yes, but this is a convention

It seems like conventions are hindering us

No, that's not a convention

It's not a convention, it's wrong.

Ah ok, i see

right

It is fundamentally why we think about limits in the first place

So, a cdf determines prob. measure because we can subtract the cdf at certain points of the domain to get the probability of certain subsets?

Very roughly speaking, yes

@Thorgott Yes

But there should be a way, however

Why do we care about cdf?

We are usually interested in subsets (events) of the sample space, so prob. measure would be sufficient

It suffices to know the values of the measure on intervals to know it on every Borel subset. That's a hard theorem.

Because the cdf uniquely determines the probability measure, we can translate information about the cdf into information about the probability measure and vice versa.

Let $u=\ln(x)$ and $v=\ln(y).$ Is it correct to conclude that $uv=1,$ in $\log-\log,$ space is a hyperbola, but in $x,y$ space it is not a hyperbola?

The measure itself is very difficult to work with, the cdf is not.

So we usually try to make do with the cdf and it often suffices.

Why is it hard to work with the measure?

It's a map from a gigantic subset of the power set of R to [0, 1]. You cannot "enlist" all Borel subsets

Hold my beer

I have no idea what Borel subsets are

It's so much easier to think of a function R -> [0, 1] than a function from a subset of P(R) to [0, 1]

Maybe you right

@Thorgott I mean, formally my procedure is not correct, but it seems to me that the result is not that "untrue", it ends up touching the reality of the problem, although through a dirty play...

@BalarkaSen the fact that a cdf induces a measure at all is much harder than its uniqueness, to be fair

True.

Event space is a synonym for $\sigma$-algebra, in the context of probability theory?

@Shootforthemoon which is probably because a) an inverse function exists and b) the derivative doesnt vanish outside of $0$ (or at least not too much) and the derivative of the implicit function is continuous

Well, every event space should be a $\sigma$-algebra. Right?

Yes, $\sigma$-algebra is abstracting out what an event space should be. In basic probability theory one only deals with finite (or countable) probability spaces, where if the sample space is $\Omega$ the event space is all of the power set $P(\Omega)$

cause discrete measures are nice

For uncountable probability spaces it gets complicated.

What kind of stuff would have an uncountable probability space? Something like produces a real number as outcome?

Right.

Hi @Balarka

Indeed, how could the probability of infinitely many numbers be 1? Actually, it makes some sense

A uniform distribution over the unit interval, to return to that example

@Thorgott True, it is thanks to the problem itself that my unfair method adapts well, but are there examples where the same method would fail?

An example would be where your sample space is $[0, 1]$ and for any subset $A \subset [0, 1]$ you assign it's probability to simply be the "volume" of $A$. This volume should satisfy the property that volume of $[a, b] \subset [0, 1]$ is $b - a$.

But it's a fact that you cannot define such a volume for every subset of $[0, 1]$. You can define it for a fairly large subset of $P([0, 1])$ however, and that's called the Borel subsets.

@Shootforthemoon take the $\sin x/x$ example above

And this notion of volume is also known as the Lebesgue measure.

Hi @Alessandro

Ok, I think I can ignore Borel for now

(you can actually do it for more subsets but using the lebesgue sigma algebra is bad for some reason that I forgot)

Ok, this is getting too measurable :P

Probabilists never use the Lebesgue alebra, I think

You definitely want $\mathbb{R}$ as codomain with the Borel-algebra, always

Cause the preimage of a Lebesgue-measurable set need not be Lebesgue-measurable, even for continuous functions

Right

Ok, but I still didn't get one thing. Is the KL divergence defined for cdfs, probability measures, pmfs, pdfs, or what?

It's compatible with the topology

A question I'm thinking about: Is there a simple example, given $n\in\Bbb N$, of a (necessarily not f.g.) group that embeds into an $n$-generated group, but not into an $(n-1)$-generated group?

@Thorgott of course, the procedure is wrong, and as you pointed out we would obtain 0/0, but I mean the reverse. If we had lim for x approaching 0 of x and the same limit of sinx, and then we fused them in a ratio, the result is what we have in fact

@nbro Wikipedia says it's define for general probability measures, but I think for your purposes it suffices to know what it means for two given CDFs

Embedding just means injective group homomorphism here?

Yes

so the method would apply well even in that case

@BalarkaSen I've been modifying the related Wikipedia, so that to provide more derivations

like unidirectional

It's the same as being isomorphic to a subgroup

@Shootforthemoon no, you would get $0/0$

yes, ok, but the converse

What about the case $n=1$ @AlessandroCodenotti

@Alessandro You want for every $n$, right?

we determine the undetermined

So maybe you shouldn't trust that entry too much, even though I am quite familiar with the KL divergence, even though apparently not enough

@Shootforthemoon I don't think I follow what you mean

Actually, I didn't modify the Kl divergence entry

@BalarkaSen yes, ideally I'd like a family of related groups, one for every $n$

Sorry, my mistake. I modified another entry

@Alessandro So countably generated groups are out of the window

@Thorgott that's easy, $\Bbb Z$ embeds into a group with $1$ generator but not into one with zero

(For $n = 2$, of course $F_2$ is an example of such a group)

The Wikipedia says "For discrete probability distributions P and Q defined on the same probability space"

Then P and Q cannot be the probability measure

Shouldn't free groups work for all $n$

@BalarkaSen yes, by the HNN result from yesterday

Because the probability space contains the probability measure

@Thorgott $F_n$ embeds in $F_2$ for all $n$

So no

@Alessandro Sounds scary

@Thorgott $F_n$ (even $F_\infty$) embeds into $F_2$ for every $n$

So? What's the KL divergence between?

It cannot be between probability measures, according to that introductory sentence

Ok, nevermind then :p

KL divergence is between probability measures

@BalarkaSen I'll probably ask on MSE later, I know next to nothing about groups so badly not f.g.

@Thorgott But then what does "For discrete probability distributions P and Q defined on the same probability space", at the related Wikipedia entry en.wikipedia.org/wiki/Kullback%E2%80%93Leibler_divergence (section definition), mean?

@Thorgott Can I open a private discussion in order not to mess up this chat?

That you have discrete probability distributions $P$ and $Q$ on the same probability space

@Shootforthemoon sure

Thanks

@Thorgott How come a probability distribution (or measure) be defined on a probability space, which contains the probability distribution (or measure)? You're not using the terms consistently now!

@nbro i think you can replace 'probability space' with 'measurable space' - ie the underlying probability measure is irrelevant

Wait but an uncountably generated group cannot embed into a f.g. one, it's too big

@loch You're saying that two probability measures are defined for the same measurable space?

Yeah

then you're looking at two probability measures defined on your measurable space, and KL divergence gives you a number

The space does not contain the distribution.

@Thorgott A measure space contains a sample space, an event space and a probability measure (or distribution). This is the definition of the probability space!

So either a group embeds into a $2$-generated one, or it doesn't embed into a f.g. group at all? That's weird

Seems like it

Crazy

@loch But the Wikipedia entry says "two distributions defined on the same PROBABILITY SPACE". A probability space is a measure space (a triple that also contains a probability measure), not a measurable space (a tuple that contains the sample and event spaces, WITHOUT the probability measure)

Then you ask why people get confused. Hm. It's not the fault of the learner. It's the fault of the writer or teacher

Ok, let's assume that the KL is defined for probability measures

What does it mean that we take the expectation of a probability measure?

There's no inconsistency here. The problem is you don't really understand what a measure is.

What?

haha

Do I need to repeat?

"two distributions defined on the same PROBABILITY SPACE"

Yes, what is the problem?

A probability space is composed of a sample space, an event space and a probability measure. Right?

Yes.

A probability distribution is a synonym for probability measure. Right?

It can be used in different contexts. It can also mean a random variable.

Ok, but in the context of the KL divergence

Right?

They mean $P$ and $Q$ are random variables on the probability space $\mathcal{X}$.

@nbro Your complaint is that the underlying measure doesn't show up in the definition and I agree! I don't think it's relevant. You take two probability measures on the space, and you get a number out of it.

So, what the heck does "two probability measures defined on the same PROBABILITY SPACE" mean???

No, they mean random variables. $P, Q : \mathcal{X} \to \Bbb R$, two discrete random variables.

@BalarkaSen What the heck is X here?

A probability space $(X, \mathcal{A}, \mu)$.

Random variables are functions on probability spaces.

A random variable goes from a probability space?

Yes

So, a distribution is not a synonym for probability measure

But @Thorgott told it is

It can mean different things in different contexts, is my point.

So, who the hell is wrong?

The fault is of the teacher

They are synonymous, the issue is that the article refers to random variables by their distributions

Nobody is. You just need to read a textbook on probability

Which is slightly sloppy, but not an issue when you understand the topic

It usually is clear from context what it means in what context

No, it's not clear

Definitions should be definitions

not relative

I'm sorry, but you are in no position to criticize that when you don't understand the foundations

What sense does "They are synonymous, the issue is that the article refers to random variables by their distributions" make?

Yes, they are apples, but the article refers to it as oranges

The fault is of the teacher

Just go through the first few chapters of a standard probability textbook, say Grimmett-Stirzaker. It really isn't hard to pick up all these.

Yes, I don't know everything

But the fault is of the teacher

Why are you acting like a broken record now

You've been told from the beginning to go read a book on the subject to learn it properly

You say "yes, X=Y", then you say "Article Z says that X=W". Then you say that I don't understand the foundations

WTF?!

Don't tell me what to do

I do what I want

lol

No, it's your fault

You should have said, the definition depends on the context

lol

As I've always thougth

It doesn't, the article is slightly imprecise

lol

@Thorgott So, you disagree with @BalarkaSen?

Because he said that the definition depends on the context

Thorgott was right, the point is, I don't think you understand how a random variable induces a measure in the first place.

Actually, you also disagree with yourself. You said the same thing above

@BalarkaSen Yes, I understand that

No, I said the same thing above

You said it above

Ok, tell me

"They are synonymous, the issue is that the article refers to random variables by their distributions"

I have a random variable $X : (\Omega, \mathcal{A}, \Bbb P) \to \Bbb R$. What is the measure induced by $X$, actually?

Here I am reaffirming that they are synonymous and also express that I think the article is being imprecise

P

False.

Well, then this is contradictory to what @Thorgott said

The measure is defined on $\Bbb R$, which takes a (Borel) subset $A \subset \Bbb R$, and spits out $\Bbb P(X^{-1}(A))$.

@BalarkaSen and @Thorgott learned different mathematics

That's why you can say it's a distribution on $\Bbb R$.

No, that doesn't contradict anything I said

Balarka and I both have a proper understanding of the foundations

So, what is P?

It doesn't contradict Thorgott, we both tried to not spoonfeed you a first course in probability.

Instead of looking for contradictions, you should try to understand what he is saying

And we're tired of trying, so I'm putting you on ignore

You don't get anything out of trying to blame either of us

What is P?

You asked "What is the measure induced by X, actually?". I answered "It is P". You said "It's false"

loch already said this yesterday, but you could've spent these hours reading some chapters of an introductory book to probability theory and learned much more

What is P?

$\mathbb{P}$ is NOT a probability measure according to @BalarkaSen but it IS a probability measure according to @Thorgott, then the fault is of @nbro, who doesn't understand the foundations.

But "induced by" is not precisely defined

You mean what you want it to mean

Then you say it is my fault and that I blame you for no reason

@loch What is the difference?

the measure induced by $X$ here is now a measure on $\mathbb{R}$

@loch Note that @Thorgott said that every r.v. induces a probability measure

whereas $\mathbb{P}$ is a measure on your original sample space

yes

@loch So, how can you blame me, if I don't know your personal definition of an expression?

@loch Ok, but how is it different from $\mathbb{P}$?

because if you actually read a book, you'll likely be able to tell what people mean! and it's much easier that way

well they are measures on different spaces

So, there are two types of measures?

Apparently, there's a measure induced by a r.v. X and then there is P, which is also a measure

You should be trying to understand what people are telling you, rather than pointing fingers, nobody cares whose fault it is anyway, especially since nobody here has to invest their time in explaining things, but are choosing to do so

Well, I am also choosing to ask

@nbro the measure induced by X is the pushforward measure of P, that's how they're related

Right, so there are 2 different measures, when we talk about r.v.s

@AlessandroCodenotti What do you mean by "pushforward"?

You should learn measure theory before measure theoretic probability

I didn't even want to learn measure-theoretic probability. I just wanted to understand probability

Then you should read a book going with the classical approach rather than the measure theoretic one

Well, I read books in the past, but I forgot many things

I thought that the expression "probability distribution" is ambiguous and not well defined

However, you guys say it is a synonym for probability measure

Now, which probability measure are we talking about?

@Thorgott Do you know anything about concentration inequalities by chance

Then you say it can also mean a random variable, because, from the context, you understand it means a random variable

Then the fault is mine!

Very well!

@nbro you could say that (but they live on different spaces!!) - but one makes sense even before mentioning anything about your random variable (your original probability measure on your sample space), while the other (the induced measure on $\mathbb{R}$) only makes sense after you have define your random variable

@loch Ok, thanks for clarifying this

I thought @Thorgott had said that a r.v. X actually induces P, which you say makes sense even before defining X

I only know the standard ones, I'm afraid @BalarkaSen

I know nothing beyond basic Chernoff. I was looking these notes, and I was wondering if you know a reference for Janson-Suen inequality in page 49

Apparently it gives exponential concentration around expectation for the expected number of triangles on a random graph

Whereas if I try by hand and estimate the covariance terms in the proof of weak law of large numbers, say, I get polynomial concentration

Something like $O(1/n^2)$

I don't even know how to begin to use Chernoff for non-iid things. As I recall the basic idea is you use Chebyshev for $e^{\lambda X}$ instead of $X$, and then since $X = \sum X_i$ is sum of iid things, $e^{\lambda X} = \prod e^{\lambda X_i}$ and the expectation also breaks up into products by independence.

Is there a noob estimate for how far the expectation of products is away from product of expectations in terms of the cross-covariances?

So, is the KL divergence defined for r.v.s or probability measures $\mathbb{P}$, or something else?

Hmm, this seems interesting, but I haven't seen this before

Chapter $5.3$ is very close to the topic one of my friends wrote his Bachelor's thesis on

Bummer. Thanks for having a look though

Oh interesting

I'll ask him whether he knows that inequality

Thanks!

Oh, so this inequality even helps studying the case of induced subgraphs

Consider the set of all continuous functions $f$ on the circle, such that $f(p)=f(q)$ for some fixed $p,q\in S^1$. We can think of a continuous function on the circle as a loop $\iota:S^1\to \Bbb R\times I$ such that $\iota\cong \gamma$ (homotopic) for $\gamma$ a generator for $\pi_1(\Bbb R\times I)\cong \Bbb Z$. So these just additionally satisfy $f(p)=f(q)$. Can we say anything interesting about the collection of such maps?

Btw, only thinking about this because of this answer: mathoverflow.net/a/45218

@TedE what is $I$?

@TedE They're the same as functions $f : S^1 \vee S^1 \to \Bbb R$, no?

@loch The interval $I=[0,1]$. I'm just thinking visually about what a continuous function with real values is, and you can think of the values as giving you the height on the cylinder at that point of the circle, so the zero function is just the circle itself.

so you meant $S^1 \times I$ then lol

Oh lol, sorry, I miswrote that

mb

(The reason for my typo) It's because on paper I'm writing $\Bbb R\times I$, and then the functions just have to agree on the endpoints, (obviously I'm really drawing $I\times I$, since my paper has finite dimensions - but I also have trouble drawing a function as a loop on $I\times S^1$))

Does it make any sense to try to understand kan's ex-infinity functor on it's own (i.e. just from the definition of a simplicial set and in a combinatorical way and maybe a bit algebraic) without having the general picture of fibrant replacement and so on?

I am not sure if the question makes sense but I wanted to ask if the general picture is mandatory (for example for dold-kan correspondence, there exists also a model categorical/infinity categorical perspective as far as I understood, but it's not that mandatory, in the sense that, you can still motivate and understand the result "locally")

@BalarkaSen I'm not seeing it

It passes through the quotient $S^1/p \sim q$ by universal property

That's a wedge of circles

As sets, $C(S^1 \vee S^1)$ is limit of the diagram $C(S^1) \to C(*) \leftarrow C(S^1)$ where the first is induced from the inclusion $\{*\} \to S^1$ as $p$ and as $q$ on the other factor.

Well, as topological spaces as well, I suppose, under the compact-open topology.

And the functor $C(-)$ commutes with limits (or atleast pullbacks I think you're claiming)?

As an aside, if you're giving the compact-open topology, then $C(-)$ is an endofunctor on Top?

Yes to second, and I think so to first with careful assumptions.

@loch Ok, thanks. Meanwhile, I've asked a formal question on the site: math.stackexchange.com/q/3483930/168764.

Wait, is $C(-)$ just $\text{Hom}_{\text{top}}(-,(\Bbb R,\tau_{\text{euc}}))$ (prior to giving compact-open topology)?

yes, also known as the set of continuous functions :p

I know, but I mean that representing it like this, you get lots of categorical properties (like it preserves colimits of topological spaces)

once you impose the topology the abstract nonsense gets thrown out of the window

you have to check everything by hand

If $A, X, Y$ are CGHaus and $A \to X, Y$ are closed inclusions, $C(X \cup_A Y)$ should be homemorphic to $C(X) \times_{C(A)} C(Y)$.

I am not sure

Well that looks like a colimit notation, so that's true since our hom functor preserves colimits

Not in sets dude

in TOP

In top

Why? Hom functor preserves colimits as sets

What do you mean?

Hom(X, Y) is nothing but a set a-priori

$\text{Hom}:\mathcal{C}^{op}\times \mathcal{C}\to \text{Set}$ preserves limits in both arguments

Yes, so as sets

I said homeomorphic, after imposing compact open topology

Oh right, I ignored the homeomorphic, my bad

I'm not wearing my reading monocle

I think it's fine

$C(X) \times C(Y)$ is naturally homeomorphic to $C(X \sqcup Y)$

Hello the math

Also, some care has to be taken to make the categorical nonsense work, like restricting to ordinal-small stuff, but I'll ignore that

I have a question on the supersymmetry

And $C(X) \times_{C(A)} C(Y)$ gets subspace topology from $C(X) \times C(Y)$, so we're looking at functions on $X \sqcup Y$ which restricts to the same functions along $A \to X, Y$.

As far as I'm aware, supersymmetry as far as math goes acts on super vector spaces, which are $\mathbb{Z}_2$-graded vector spaces

But this requires them to be vector spaces on the same fields

But on the other hand, physically speaking, those are usually scalar fields (which are $\approx \mathbb{R}$) and spinor fields ($\approx \mathbb{C}^2$)

Yea, I've only seen supersymm stuff with $\Bbb{Z}_{2}$ graded things