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

**1****st**** 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.

**2****nd**** 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.