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Let $f$ be a function from $B$ to $C$ and $g$ be a function from $A$ to $B$, such that $f \circ g$ is one-to-one.

a) Prove that $g$ is one-to-one

b) Prove that if $g$ is onto, then $f$ is one-to-one

I'm needing some help with this problem. For a, could I assume that $f$ is one-to-one in order to prove that $g$ is one-to-one. Such that:

Assume $f(w) = f(z) ⇒ w = z$ and $f \circ g$ is $1-1$ where $f \circ g(x) = f \circ g(y)$.

$f(g(x)) = f(g(y))$ from $f \circ g(x) = f \circ g(y)$

and based off $f(w) = f(z) ⇒ w = z$, we can say $g(x) = g(y)$

then $x = y$

therefore g is one-to-one

For part b, this is as far as I understand to go:

Assume $g$ is onto and $f \circ g$ is one-to-one

Then $g(x) = g(y)$ and $g(w) = g(y)$

$x = y$ and $w = y$

  • 1
    For a) you can't assume that $f$ is injective. – 2017-02-25
  • 1
    See http://math.stackexchange.com/questions/2063983/composite-set-question-with-functions – 2017-02-25

1 Answers 1

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(a) Let $a_1,a_2 \in A$ such that $g(a_1)=g(a_2)$. We wish to show $a_1=a_2$.

\begin{align} g(a_1)&=g(a_2) \\ (f\circ g)(a_1)&=(f \circ g)(a_2) \\ a_1&=a_2 \quad(f \circ g\text{ is injective}) \end{align}

(b) Let $b_1,b_2 \in B$ such that $f(b_1)=f(b_2)$. We wish to show $b_1=b_2$.

Since $g:A\to B$ is surjective, $\exists a_1,a_2\in A$ such that $g(a_1) = b_1$ and $g(a_2) = b_2$.

\begin{align} f(b_1)&=f(b_2) \\ f(g(a_1))&=f(g(a_2)) \\ (f\circ g)(a_1)&=(f \circ g)(a_2) \\ a_1&=a_2 \quad(f \circ g\text{ is injective}) \\ g(a_1)&=g(a_2) \\ b_1&=b_2 \end{align}