MHB Compose Functions: True/False?

[lemonthree]

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.

 

[Evgeny.Makarov]

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.

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.

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".)

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.