MHB Compose Functions: True/False?

  • Thread starter Thread starter lemonthree
  • Start date Start date
  • Tags Tags
    Functions
AI Thread Summary
The discussion centers on the properties of function composition, specifically regarding injective, surjective, and bijective functions. It is established that if both f and g are injective, surjective, or bijective, then their composition g∘f retains those properties. However, it is clarified that if g∘f is bijective, it does not necessarily imply that either f or g must also be bijective. The main point of contention is whether f∘g equals g∘f when A, B, and C are equal sets; the consensus is that equality of sets does not guarantee the equality of function compositions. The conclusion emphasizes that function equality requires matching outputs for all inputs, not just matching domains and codomains.
lemonthree
Messages
47
Reaction score
0
Which of the following are true? Select all options. Assume that f:A→B and g:B→C.

If f and g are injective, then so is g∘f
If f and g are surjective, then so is g∘f
If f and g are bijective, then so is g∘f
If g∘f is bijective, then so are both f and g
If g∘f is bijective, then so is either f or g
If A=B=C, then f∘g=g∘f.

The first 3 options are true while the next 2 options are false. The only question I have here is regarding the last option "If A=B=C, then f∘g=g∘f."

Why can't it be true? If A = B = C, and they are all equal sets, we know that f:A→B and g:B→C.
So in essence, isn't f∘g=g∘f?
f∘g= f(g(x)) = f(C) = f(A) = B
While g∘f = g(f(x)) = g(B) = C
But A = B = C so they are all equal.
Range = codomain = domain anyway.
 
Mathematics news on Phys.org
lemonthree said:
The only question I have here is regarding the last option "If A=B=C, then f∘g=g∘f."
Then it makes sense to post only that question.
lemonthree said:
Why can't it be true?
The question is not whether $f\circ g=g\circ f$ can be true, it's whether it must be true under the given assumptions.
lemonthree said:
If A = B = C, and they are all equal sets, we know that f:A→B and g:B→C.
More than that, we are given that $f:A\to B$ and $g:B\to C$ even if $A=B=C$ does not hold.

(A relevant joke. A tourist asks a local: "If I go down this street, will there be a railway station? The local replies, "The station will be there even if you don't go down that street".)

lemonthree said:
f∘g= f(g(x)) = f(C)
$f(g(x))$ cannot equal $f(C)$ because the former is an element of the codomain of $f$ while $f(C)$, if defined, is a subset of the codomain of $f$. In any case, equality of two functions $h_1$ and $h_2$ means that $h_1(x)=h_2(x)$ for every $x$ in the domain of $h_1$ and $h_2$ and not just that the domain and codomain of $h_1$ and $h_2$ are the same.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Thread 'Unit Circle Double Angle Derivations'
Here I made a terrible mistake of assuming this to be an equilateral triangle and set 2sinx=1 => x=pi/6. Although this did derive the double angle formulas it also led into a terrible mess trying to find all the combinations of sides. I must have been tired and just assumed 6x=180 and 2sinx=1. By that time, I was so mindset that I nearly scolded a person for even saying 90-x. I wonder if this is a case of biased observation that seeks to dis credit me like Jesus of Nazareth since in reality...
Thread 'Imaginary Pythagoras'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...
Back
Top