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a) If $f: E \to F$ and $g: F \to E \ $ are functions such that $g \circ f$ is injective, then $f$ is injective

b) If $f: E \to F$ and $g: F \to E \ $ are functions such that $g \circ f$ is surjective, then $g$ is surjective.

Proof:

a) Assume $g \circ f$ is injective but $f$ is not injective. Then there exist $x_1, x_2 \in E$ such that $x_1 \neq x_2$ but $f(x_1) = f(x_2)$. Then $g(f(x_1)) = g(f(x_2))$. Since $g\circ f$ is injective, $x_1 = x_2$, a contradiction! Therefore $f$ is injective.

b) Assume $g \circ f$ is surjective but $g$ is not surjective. Then there exists $e_1 \in E$ such that $e_1 \neq g(z)$ for all $z \in f$. Since $g\circ f$ is surjective, there exists $x_1 \in E$ such that $g(f(x_1)) = e_1$. But this is a contradiction to the hypothesis that $g$ is not surjective, since $f(x_1) = y_1 \in F$! Therefore $g$ is surjective.

Is everything here correct or did I make a mistake somewhere? If anyone has a cleaner proof I'd appreciate it as well, since I have still not convinced myself of the validity of the one I wrote above.

Community
  • 1
Matheus Andrade
  • 6,998

2 Answers2

7

It is okay, but can be done more directly.

Let $f(x_1)=f(x_2)$.

Then $(g\circ f)(x_1)=g(f(x_1))=g(f(x_2))=(g\circ f)(x_2)$ so the injectivity of $g\circ f$ tells us that $x_1=x_2$.

Proved is now that $f$ is injective.

Let $e\in E$.

Since $g\circ f$ is surjective we have $g(f(x))=(g\circ f)(x)=e$ for some $x$.

That means that $g$ must be surjective.

drhab
  • 153,781
2

1) Injectivity has been dealt with.

2) Surjectivity:

$f: E \rightarrow F$; $g: F \rightarrow E.$

Show that:

If $g \circ f$ is surjective then $g$ is surjective:

$g\circ f$ : $E \rightarrow E.$

Let $e_2 \in E$, then,

since $g \circ f$ surjective, there is an $e_1 \in E$ such that

$(g\circ f)(e_1)=g(f(e_1))=e_2$, i.e.

there is an element $z= f(e_1) \in F$ with

$g(z)=e_2$, hence surjective.

Peter Szilas
  • 21,123