Thread: Bijective Proof

Hey there I was just hoping some of you could check out my proof and see if you could improve it any way. My professor wanted this to be much more of a "rigorous" proof than the one before you so maybe you can help me out. Thank you guys in advance!!!


Question:

Let g be a function from x to y and let f by a function from y to z. Prove or disprove if g and f are both bijective, then f ◦ g is bijective


Direct Proof
Can be broken down into two statements because bijective implies both injectiveness and subjectiveness


1st Part – If f and g are injective then f ◦ g is injective
Because f and g are both injective this creates a situation in which any instance of f that f(x1) = f(x2) this implies that x1 = x2 AND for any instance of g that g(y1) = g(y2) this implies y1 = y2.


f(x1) = f(x2) -- > x1 = x2
*f(g(y1)) = f(g(y2)) --> g(y1) = g(y2)


Because of this definition no matter which g(y) is put into f(g(y)) it will be a unique value which implies that f ◦ g is injective.


2nd Part – If f and g are subjective then f ◦ g is subjective
Because f and g are subjective this implies that for...


f there is Ay Ex(f(x) = y)
g there is Az Ex(g(y) = z)
*f there is Ag Ex(f(g(y)) = y)


Because f and g are both subjective this creates a situation in which f's domain [x] maps to its entire codomain [y] and g's domain [f's codomain/Y] maps entirely to its codomain[z]. This creates a situation in which f ◦ g is subjective because any output of g can be mapped to f's codomain.


* Direct Proof – Due to parts 1 and 2 above it is shown that if g and f are both bijective, then f ◦ g is bijective.

Originally Posted by Eriktg

Hey there I was just hoping some of you could check out my proof and see if you could improve it any way. My professor wanted this to be much more of a "rigorous" proof than the one before you so maybe you can help me out. Thank you guys in advance!!!


Question:

Let g be a function from x to y and let f by a function from y to z. Prove or disprove if g and f are both bijective, then f ◦ g is bijective


Direct Proof
Can be broken down into two statements because bijective implies both injectiveness and subjectiveness


1st Part – If f and g are injective then f ◦ g is injective
Because f and g are both injective this creates a situation in which any instance of f that f(x1) = f(x2) this implies that x1 = x2 AND for any instance of g that g(y1) = g(y2) this implies y1 = y2.


f(x1) = f(x2) -- > x1 = x2
*f(g(y1)) = f(g(y2)) --> g(y1) = g(y2)


Because of this definition no matter which g(y) is put into f(g(y)) it will be a unique value which implies that f ◦ g is injective.


2nd Part – If f and g are subjective then f ◦ g is subjective
Because f and g are subjective this implies that for...


f there is Ay Ex(f(x) = y)
g there is Az Ex(g(y) = z)
*f there is Ag Ex(f(g(y)) = y)


Because f and g are both subjective this creates a situation in which f's domain [x] maps to its entire codomain [y] and g's domain [f's codomain/Y] maps entirely to its codomain[z]. This creates a situation in which f ◦ g is subjective because any output of g can be mapped to f's codomain.


* Direct Proof – Due to parts 1 and 2 above it is shown that if g and f are both bijective, then f ◦ g is bijective.

.

Let: g:A---->B, f: B----->C.

You want to prove :g,f surjections ====> fog surjection.

The argument goes like this:

Let zεC,THEN since

1) f,is surjection ,then for all z ,zεC, THERE exists yεB SUCH that f(y)=z.

Thus ,there exists yεΒ and f(y)=z............................................ ........1

2) g,is surjection ,then for all yεB THERE exists ,xεA SUCH that g(x)= y.

Thus ,there exists xεA AND g(x)=y............................................ .......2

By substituting (2) into (1) we have: f(g(x)) = z


Hence for all zεC THERE exists xεA SUCH THAT f(g(x))= z ,by substituting (2) into (1)
Thus fog is surjective