Are f and g Injective and Surjective if g\circf is Injective or Surjective?

[Ka Yan]

Could anybody help me check whether my judgements ture or false? (MJ = My Judgement)

Suppose f maps A into B, and g maps B into C

1. If f and g are injective, then g[tex]\circ[/tex]f is injective;
(MJ)but that when g[tex]\circ [/tex]f is injective, the injectivity of f and g are unsure.

2. If f and g are surjective, then g[tex]\circ[/tex]f is surjective;
(MJ)and that when g[tex]\circ[/tex]f is surjective, f and g are both surjective.

Thx!

 

[HallsofIvy]

What reasons do you have for those judgements? Can you think up a simple example where f and g are not injective but g[itex]\circ[/itex]f is? Try the case where A, B, and C have only a few members.

 

[Architeuthis Dux]

1. True. If f and g are injective, then for any elements a and b in A, if f(a) = f(b), then a = b. Similarly, for any elements b and c in B, if g(b) = g(c), then b = c. This means that if g\circf(a) = g\circf(b), then a = b, making g\circf injective.

2. True. If f and g are surjective, then for any element c in C, there exists an element a in A such that f(a) = b, and for any element b in B, there exists an element c in C such that g(b) = c. This means that for any element c in C, there exists an element a in A such that g\circf(a) = c, making g\circf surjective.