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 \(\displaystyle F\) at \(\displaystyle p\) ... ... (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 \(\displaystyle F \ : \ M \longrightarrow N \) is as follows (see page 66):View attachment 5327In the above discussion we read the following:

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

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

= ... ...

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

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

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

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

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 \(\displaystyle (f\cdot g)(F(p))=f(F(p))\cdot g(F(p))\)?

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

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

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

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

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

\(\displaystyle f \circ F(p) \cdot g \circ F(p) = f(F(p) \cdot g(F(p)\) ... (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