Injective and surjective proof help I am doing my homework, but with classes being cancelled and online, it is difficult for me to learn and understand these questions, I have pasted most of them below and attempted to answer to the best of my ability. Any tips or corrections would be greatly appreciated!

*

*Prove that if $f : A \longrightarrow B$, $g:B \longrightarrow C$, and $g \circ f:A \longrightarrow C$ is a 1-to-1 surjection, then $f$ is 1-to-1 and $g$ is a surjection.

Proof. Assume $g \circ f$ is  a  1-to-1  surjection.   Assume $x,y \in A$ such that $f(x) =f(y)$.  Then $(g \circ f)(x) = g(f(y) = (g \circ f)(y)$.  But since $g \circ f$ is 1-to-1, $x=y$.  Therefore, $f$ is 1-to-1.  Next, let $z \in C$.  Then, since $g \circ f$, there exists $x \in A$ such that $(g \circ f)(x) = (g(f(x)) =z$. Therefore, if $y=f(x) \in B$, then $g(y) =z$.  Thus, $g$ is surjective
(I have finished this one but am not very confident with it)


*Prove  that  if $f: A \longrightarrow B , g : B \longrightarrow C , g \circ f : A \longrightarrow C$ is  a  1-to-1 surjection, and $g$ is 1-to-1, then $f$ is a surjection.

Proof: Assume $g \circ f$ is a 1-to-1 surjection.  Let $y \in B$.  Since $g \circ f$ is surjective, there exists an $x \in A$ such that $g \circ f(x) = g(y)$.  That is, $g(f(x)) =g(y)$.....
(Haven't finished this one)


*Prove  that  if $f:A \longrightarrow B$ , $g: B \longrightarrow C$, $g \circ f : A \longrightarrow C$ is  a  1-to-1 surjection, and $f$ is a surjection, then $g$ is 1-to-1.
Proof. Assume $g \circ f$ is a 1-to-1 surjection and $f$ is a surjection.  Let $x,y \in B$ such that $g(x) =g(y)$.  Than $(g \circ f)(x) =g(y) = (g \circ f)(y)$. But since, $g \circ f$ is 1-to-1 , $x=y$.  Thus, $g$ is 1-to-1.
(Tried to base it off the first question, but not positive)
Apologies if some of the format is off! It didn't copy perfectly from overleaf but I did my best to fix it, I may have missed a few things.
 A: Your solution to $1$ is correct. 
For $2$, since you are told that $g$ is $1$-to-$1$, your $g(f(x)) = g(y)$ gives $f(x) = y$. Accordingly, for every $y \in B$, you have found an $x \in A$ that is mapped by $f$ to $y$, so $f$ is surjective as required.
Your solution to $3$ is not correct, since you have written $(g \circ f)(x) = g(y)$, but there is no reason to think $y = f(x)$. A correct solution would be as follows: take $x, y \in B$ with $g(x) = g(y)$. Since $f: A \to B$ is a surjection, there are $t, u \in A$ with $x = f(t)$ and $y = f(u)$. Thus $g(x) = g(y)$ becomes $gf(t) = gf(u)$. Since $g \circ f$ is $1$-to-$1$, this gives $t = u$, so $x = f(t) = f(u) = y$, meaning that $g$ is $1$-to-$1$ as required.
For $4$, try $f(x) = \arctan x$ and $g(x) = \tan x$, so that $gf(x) = x$. Both of these are well-defined for all $x \in \mathbb{R}$, and return real outputs, so are functions from $\mathbb{R} \to \mathbb{R}$. $f$ is not a surjection since $\arctan x$ always returns a value in $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$, so for example there is no $x \in \mathbb{R}$ with $\arctan x = 2$. $g$ is not $1$-to-$1$ as $\tan x$ is periodic, so e.g. $\tan 0 = \tan \pi = 0$. But clearly $g \circ f$, being the identity function $x$, is clearly a bijection (or a "$1$-to-$1$ surjection" as you put it).
