MHB Distributing the Product of Functions over Composition of Functions

[Math Amateur]

I am reading John M. Lee's book: Introduction to Smooth Manifolds ...

I am focused on Chapter 3: Tangent Vectors ...

I need some help in fully understanding Lee's definition and conversation on pushforwards of $$F$$ at $$p$$ ... ... (see Lee's conversation/discussion posted below ... ... )

Although the book is on differential topology, my question is essentially algebraic ...

Lee's definition and discussion of pushforwards of $$F \ : \ M \longrightarrow N $$ is as follows (see page 66):View attachment 5327In the above discussion we read the following:

" ... ... $$(F_*X)(fg) = X((fg) \circ F)$$

= $$X((f \circ F) (g \circ F) ) $$

= ... ...

... ... "The above working by Lee implies that

$$(fg) \circ F = (f \circ F) (g \circ F)$$

but why is this true ... ?

Can someone help ...

Peter*** EDIT ***

Just thinking ... if R was an algebra of smooth functions with operations of addition and multiplication of functions and a 'multiplication' that was composition of functions ... then would multiplication of functions distribute over composition ... just a thought ... but I suspect a bit muddled ...

Maybe if one could justify that functions under product and composition of functions ... but you would have to assume the required distributivity anyway ...

Must be a more direct way ...

 

[GJA]

Hi Peter,

I am reading John M. Lee's book: Introduction to Smooth Manifolds ...

The above working by Lee implies that

$$(fg) \circ F = (f \circ F) (g \circ F)$$

but why is this true ... ?

This is the result of the following computation: $((f\cdot g)\circ F)(p) = (f\cdot g)(F(p))=f(F(p))\cdot g(F(p)) = (f\circ F)(p)\cdot (g\circ F)(p)$. Since $p$ was arbitrary, $(f\cdot g)\circ F = (f\circ F)\cdot (g\circ F).$ Hope that helps! Let me know if there's still confusion.

 

[Math Amateur]

Hi Peter,
This is the result of the following computation: $((f\cdot g)\circ F)(p) = (f\cdot g)(F(p))=f(F(p))\cdot g(F(p)) = (f\circ F)(p)\cdot (g\circ F)(p)$. Since $p$ was arbitrary, $(f\cdot g)\circ F = (f\circ F)\cdot (g\circ F).$ Hope that helps! Let me know if there's still confusion.

Thanks GJA ... appreciate your help ...

... but ... I need a further clarification ...Why, exactly is $$(f\cdot g)(F(p))=f(F(p))\cdot g(F(p))$$?

Can you explain ...?

Thanks again ...

Peter

 

[GJA]

Hi Peter,

I was trying to be explicit about multiplication using the "dot" versus the composition using the "circle."

The definition of function multiplication is given by $(f\cdot g)(x) := f(x)\cdot g(x).$

The definition of function composition is given by $(f\circ F)(x) := f(F(x)).$

Combining these two definitions gives the computation I presented in the previous post. Hope that helps!

 

[Math Amateur]

Hi Peter,

I was trying to be explicit about multiplication using the "dot" versus the composition using the "circle."

The definition of function multiplication is given by $(f\cdot g)(x) := f(x)\cdot g(x).$

The definition of function composition is given by $(f\circ F)(x) := f(F(x)).$

Combining these two definitions gives the computation I presented in the previous post. Hope that helps!

Hi GJA ... yes, understand those two definitions ...

... but ... is the equation

$$(f\cdot g)(F(p))=f(F(p))\cdot g(F(p))$$

simply the result of applying the two definitions that you mention above ... ?It looks to me as if the $$\cdot$$ is distributed over the $$\circ$$ as follows:

$$(f\cdot g)(F(p)) = ( f \cdot g ) \circ F (p)$$ ... ... ... ... (1) (just definition!/notation)then the "distribute $$\cdot$$ over $$\circ$$" step:

$$( f \cdot g ) \circ F (p) = f \circ F(p) \cdot g \circ F(p)$$ ... ... ... (2)and then we have

$$f \circ F(p) \cdot g \circ F(p) = f(F(p) \cdot g(F(p)$$ ... (3) (just definition/notation)
I understand equations (1) and (3) because they simply rest on the definitions you mention ... but I still fail to see why step (2) holds ... it does not seem to me to be just a matter of the definitions you mention ...

Can you help/clarify further ... ...

Peter

 

[GJA]

Hi Peter,

Let's try to simplify the notation by setting $x$ equal to the point $F(p)\in N$; i.e. let $x=F(p)$. Then we can write

$((f\cdot g)\circ F)(p) = (f\cdot g)(F(p)) = (f\cdot g)(x) = f(x)\cdot g(x)=f(F(p))\cdot g(F(p)) = (f\circ F)(p)\cdot (g\circ F)(p).$

The first equality follows from the definition of function composition, the second by our definition of $x$, the third by the definition of function multiplication, the fourth by our definition of $x$, the fifth by the definition of function composition.

Note that all of the justifications above either involve a definition of function combinations (multiplication or composition), or the use of $x=F(p).$ So, had we not taken the extra step of setting $x=F(p)$, then the only things needed to justify the equality you're after is the definitions of function multiplication and function composition.

Does this help clear things up at all?

Edit: $F(p)$ is an element of $N$, not $\mathbb{R}$ as I originally said.

 

[Math Amateur]

Hi Peter,

Let's try to simplify the notation by setting $x$ equal to the point $F(p)\in N$; i.e. let $x=F(p)$. Then we can write

$((f\cdot g)\circ F)(p) = (f\cdot g)(F(p)) = (f\cdot g)(x) = f(x)\cdot g(x)=f(F(p))\cdot g(F(p)) = (f\circ F)(p)\cdot (g\circ F)(p).$

The first equality follows from the definition of function composition, the second by our definition of $x$, the third by the definition of function multiplication, the fourth by our definition of $x$, the fifth by the definition of function composition.

Note that all of the justifications above either involve a definition of function combinations (multiplication or composition), or the use of $x=F(p).$ So, had we not taken the extra step of setting $x=F(p)$, then the only things needed to justify the equality you're after is the definitions of function multiplication and function composition.

Does this help clear things up at all?

Edit: $F(p)$ is an element of $N$, not $\mathbb{R}$ as I originally said.

Thanks GJA ... most helpful of you ...

Everything made clear!

Peter