If $ƒ$ and $ƒ \circ g$ are one-to-one, does it follow that $g$ is one-to-one?

Question

I'm having some trouble understanding the steps required to prove this. Let $P(x)$ represent x is one-to-one.

Premises:

1. $P\bigl(f(x)\bigr) =\bigl(f(x_1) = f(x_2) \to x_1=x_2\bigr)$
2. $P\bigl(f\bigl(g(x)\bigr)\bigr)=\bigl(f\bigl(g(x_1)\bigr)=f\bigl(g(x_2)\bigr) \to x_1=x_2\bigr)$

Conclusion:

3. $\bigl(P\bigl(f(x)\bigr) \land P\bigl(f\bigl(g(x)\bigr)\bigr)\bigr) \to P\bigl(g(x)\bigr)$

Proof by contraposition would require me to show that

4. $\lnot P\bigl(g(x)\bigr) \to \bigl(\lnot P\bigl(f(x)\bigr) \lor \lnot P\bigl(f\bigl(g(x)\bigr)\bigr)\bigr)$

(4) can be simplified using disjunctive syllogism with (1) to

5. $\lnot P\bigl(g(x)\bigr) \to \lnot P\bigl(f\bigl(g(x)\bigr)\bigr)$

This is where I get lost. Should I simplify

6. $P\bigl(g(x)\bigr) = \bigl(g(x_1) = g(x_2) \land x_1\neq x_2\bigr)$ to yield
7. $g(x_1) = g(x_2)$ which is consistent with $P\bigl(f(\bigl(g(x)\bigr)\bigr) = \bigl(f\bigl(g(x_1)\bigr) = f(\bigl(g(x_2)\bigr) \land x_1\neq x_2\bigr)$, therefore the conclusion is true by contraposition?

Answer 1

I don't see the point of doing logic as a substitution for thinking.

Definition of $g$ being 1-1 means that if $g(x) = y$ then $x$ is distinct in that. That is to say,$g$ is 1-1 if $g(x) = g(w) \implies x = w$.

(Alternatively, the contrapositive says $g$ is 1-1 if $x \ne w \implies g(x) \ne g(w)$.

Suppose $g(x) = g(w)$ then $f(g(x)) = f(g(w))$. But $f\circ g$ is 1-1 we were told. So that means $x = w$.

So yes, if $f\circ g$ is 1-1 then $g$ is 1-1.

NOTE!!!! We don't actually need $f$ to be 1-1$

It's quite possible for $f(v) = f(g(x))= z$ without have $v = g(x)$. But if this happens then there is no $w$ so that $g(w) = v$. (Because then $g(w) = v \ne g(x)$ but $f(g(w))=f(g(x))$ which is impossible.)

So it is possible for $f\circ g$ being 1-1 but $f$ not being 1-1. But if that happens $g$ is not surjective.

Perhaps a better way to put this.

if $g: X \to Y$ and $f: Y \to Z$ and $f\circ g: X \to Z$ and if $f\circ g$ is 1-1 then:

1) $g$ is one to one.

2) $f$ restricted to: $\widehat{f}:g(X)\to Z$ is one to one.

3) But $f$ need not be one to one if $g(X) \subsetneq Y$ and $g$ is not surjective. If this occurs and $f(w) = f(g(x))$ then $w \not \in g(X)$.

Answer 2

Let $x$ and $x'$ be distinct. If the equation $g(x) = g(x')$ holds then how could the following $(f \circ g)(x) = f(g(x)) \not = f(g(x') = (f\circ g)(x')$ hold?