How do I evaluate (hog)of?

I am confused as to how to evaluate (hog)of

Do I map elements from B->C and from C->D? Then what do I do with the f? Because f maps from A->B.

This doesn't make any sense to me.

Bump

You read function compostion from left to right.
So [math]h \circ g: B \rightarrow D, f: A \rightarrow B[/math] and hence [math](h \circ g) \circ f:A \rightarrow D[/math]

why can you just write
hogof
without the brackets

I don't understand your question man. h, g, and f are just operators. If you're referring to your picture, it looks like f maps A to B, g maps B to C, and h is not a function from C to D but it maps two elements in C to two elements in D.

I cant tell if your trolling or retarded

And before some retard tries to tell me h is a function he should remember that it's only a function C-->D if it maps every element of C to an element in D.

Because composition of functions is associative so there's no ambiguity in the order of operations.

We can easily define h to be a function, are you still in highschool? Since that's a highschool definition.
>operators
>functions
>maps
>all in one post
If you're really not in highschool, then you should know better and really know the difference between these.

thanks user

Sure you can define h to be a function, but it can only be a function on a subset of C (not C itself) so (hog)of can't be a function on A (just a subset of A).

I don't think that's the answer to OP's question, but you have people like this who are wrong in saying that the composition maps A to D.

>but it can only be a function on a subset of C

those are some big claims, user

fuck me, nvm. I never looked at OPs dumb pic

You realise that a function can have elements in its domain and not map them to anywhere right?
>I've never seen functions with gaps in them before
You're even confused by OP's question and you use words like operator, maps and functions without even caring about what they mean, yet you choose to take an arbitrary definition of function really seriously.
Are you an engineer by any chance?

>oh boy here we go

I smell the genesis of even more esoteric terminology

Thanks for the explanation OP here,

So (hog)of works like this....?

hog

g: B->C
h: C-D
f: A->D

that it?

if so then I think I understand it

I am still wondering how set A "knows" how to 'skip' sets B, C and go straight to D

No no, read the post again.
What happens is that [math]f[/math] goes from [math]A[/math] to [math]B[/math], then [math]g[/math] takes [math]B[/math] to [math]A[/math], and then finally [math]h[/math] takes [math]C[/math] to [math]D[/math].
So together, [math]h \circ g \circ g[/math] takes [math]A[/math] to [math]B[/math] to [math]C[/math] to [math]D[/math].

Sorry, this should be [math]g[/math] takes [math]B[/math] to [math]C[/math].

A function on a domain X can't have elements in X that don't map to anywhere, or else it wouldn't be a function on X by pretty much every definition. What you say would be a partial function, which is a function defined on a subset of the domain X, but is not a function on the whole domain.

How? The terminology is commonplace and distinguished all over maths. Heck, even bounded and linear next to each other doesn't necessarily mean that we're bounded on the entire domain.
You really need to be careful with what you mean when you say things.

if the parentheses are around the (hog) then why are you evaluating f first?

please just kill yourself

>A function on a domain X can't have elements in X that don't map to anywhere, or else it wouldn't be a function on X by pretty much every definition.
You're being picky about definitions yet you're not careful about them yourself. I think we're pretty much done here because this discussion with you isn't teaching either person anything.

No I'm serious. If parenthesi are around (hog) then why evaulate f before (hog) this is the source of my confusion

>I can't refute this because I'm wrong so I'll bug you about using lamens terms earlier.

I'm very sorry for using the word maps sempai please forgive me, but why isn't the exact definition of a function relevant when you're trying to evaluate one?

FUNCTION COMPOSITIONS ARE EVALUATED THE OTHER WAY AROUND. REEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE

Ok I get it

You do (hog)

which is

g: B->C
h: C->D

then f

which is A->B

and you "connect" to get A->D

??????????????
There's nothing to refute when we're talking about definitions which change depending on the field that you're talking about.
The reason I was being picky with your maps/operators/functions is precisely that reason.
OP called them functions, and they are indeed functions in most areas except highschool.
It is perfectly fine to have functions map elements of their domain to nowhere, and we very much have these in topology.
All in all, there is no "exact" definition of function that is relevant here and it most definitely isn't the intent of the question set by the OP. I understand that highschool textbooks like to define functions like that, including their "many-to-one" thing and "one-to-many" thing not being a function, but the truth is that in maths we really do not care as it places an unnecessary restriction on what we can work with.

Let us write it out and evaluate it at some [math]x \in A[/math] to perhaps help you to see what is happening. So we have [math]h(g(f(x))) = h(g(y)) = h(z)[/math], where [math]y = f(x)[/math] and [math]z = g(y) = g(f(x))[/math].
Does that help?

>OP called them functions, and they are indeed functions in most areas except highschool.
The point to this (which I forgot to type) was that you shouldn't confuse OP with terms such as operators/maps when aren't correct and could lead to more unneeded confusion since OP is having problems with something like this already.

Yes,

you first evaluate f, followed by g, followed by h

but what confuses me is this is the exact same thing as (gof)oh

I thought by definition

(hog)of would be different in the evailation....

that you would do g, followed by h, followed by f... but in your example you did f, followed by g, followed by h


that is how i thought i understood it before your post

[math](g \circ f) \circ h[/math] and [(h \circ g) \circ f[/math] are NOT the same.
[math](g \circ f) \circ h[/math] is a function that doesn't even make sense because [math]g \circ f[/math] goes from [math]A[/math] to [math]B[/math] to [math]C[/math] but [math]h[/math] goes from [math]C[/math] to [math]D[/math], so we cannot apply [math]g \circ f[/math] to [math]h[/math].

ok im so fucking confused lol

(hog)of

g: B->C

then h

C->D

then f

A->B

now how does this mean A->D

Okay, perhaps it would help if we reduced on notation. Remember also that we read function composition from left to right.
So, [math]h \circ g[/math] is a function which does [math]B \rightarrow C \rightarrow D[/math] because [math]g: B \rightarrow C[/math] and [math]h: C \rightarrow D[/math]. Okay so far?
Then, let us define a new function [math]A = h \circ g[/math].
So now we care about [math]A \circ f[/math].
So since [math]f: A \rightarrow B[/math] and [math]A: B \rightarrow D[/math], we have that [math]A \circ f[/math] does [math]A \rightarrow D[/math].

oooh i see where my error in thinking was now.

that clears it up

thank you so fucking much

You're welcome user.

this now allows me to move on with my self-studies in abstract algebra and so on. now i understand the proofs that composition is associative. people like you make this board much better

Well, it took a long time to explain but you understood it in the end so it was worth it.
Happy studying user!