# Composite function proofs

Prove that this is incorrect

1. $$g \circ f$$ is surjective, but $$g$$ is not surjective.

2. $$f$$ and $$g$$ are not injective, but $$g \circ f$$ is injective.

I know these both don't exist but, I don't really know how to formally prove them. Do I use proof by contradiction?

My attempt:

1. We know $$g \circ f$$ is surjective and we want to show that $$g$$ is surjective. Let $$y \in\mathbb R$$. since $$g \circ f$$ is surjective, there exists an $$b \in \mathbb R$$ such that $$g(f(b)) = y$$. We set $$c = f(b) \in\mathbb R,$$ then $$g(f(b))=g(c)=y$$. So $$g$$ is surjective.

Is this how you're suppose to prove the question? I feel like I just proved that $$g \circ f$$ is surjective then $$g$$ is surjective but not the question above.

1. I don't really know where to start with this one. Am I suppose to use proof by contradiction here?

For part 2, since $$f$$ is not injective, $$\exists x,y \in \mathbb{R}$$ with $$x \neq y$$ such that $$f(x) = f(y)$$. Then $$(g \circ f)(x) = (g\circ f)(y)$$.

Your proof of part 1 is correct. Since $$g\circ f$$ surjective $$\implies$$ $$g$$ surjective, there is no example of $$g$$, $$f$$ where $$g \circ f$$ is surjective, but $$g$$ is not.

Edit: Since $$f(x)=f(y)$$, we have $$(g\circ f)(x) = g(f(x)) = g(f(y)) = (g \circ f)(y).$$ Since $$x \neq y$$, it follows that $$g \circ f$$ is not injective.

• Sorry, I don't get why "with 𝑥≠𝑦 such that 𝑓(𝑥)=𝑓(𝑦). Then (𝑔∘𝑓)(𝑥)=(𝑔∘𝑓)(𝑦)". Wouldn't it be (𝑔∘𝑓)(𝑥)≠(𝑔∘𝑓)(𝑦) if 𝑥≠𝑦? – bob Mar 20 at 2:04
• @bob. If $f(x)$ and $f(y)$ are the same thing then $(g f)(x)=g (f(x))=g(f(y))=(g f)(y).$ – DanielWainfleet Mar 20 at 2:37

In general, it is simple to show "that such examples do not exist."

If someone asks you to show that:

examples of the set $$A$$ do not exist

then simply prove:

$$A$$ is the empty set.

Alternatively, suppose that that someone says,

Show that there are no examples of $$x$$ in $$S$$ such that $$P(x)$$, where $$P(x)$$ is some statement about $$x$$.

You are being asked to show:

not $$[\exists x \in S$$ such that $$P(x)]$$

That's very simple. Just prove:

$$\forall x \in S$$, not $$P(x)$$

If $$gf$$ is surjective then $$\Bbb R=\{(gf)(x):x\in \Bbb R\}=$$ $$=\{g(f(x)):x\in \Bbb R\}=$$ $$=\{g(y):y\in \{f(x):x\in \Bbb R\}\}\subset$$ $$\subset \{g(y):y\in \Bbb R\}\subset \Bbb R.$$ So $$\Bbb R\subset \{g(y):y\in \Bbb R\}\subset \Bbb R.$$

So $$\{g(y):y\in \Bbb R\}=\Bbb R.$$