When is and isn’t a Surjection ‘self evident.’ studying my first ever abstract maths course (how exciting), and one thing that often comes up is that when proving a surjection…sometimes it’s just given as ‘self evident,’ other times no matter how simple…we have to do some work to get marks. I want to understand where the line is, and get some pointers.
’Self evident’ example
For example: The surjective property of a map between a subgroup $H$ and a left Coset $aH$ is seen as self-evident.
Map $H \to aH, F(h) = ah$
“Take $x \in aH$, then $x = ah$, for some $h \in H$ by definition. So we have  $F(h) = x$ thus our map is surjective.”
Question: I appreciate this does seem obvious, but can anyone show me any subtle changes that might seem right, but would make this answer actually wrong. I.e. when the answer is so simple, I feel like even the smallest change would throw it off.
Non self evident example
$$F(x) = x^2 + 1$$
Here we actually have to find the inverse, so that we can show that always, $f(f^{-1}(x))$ gives us our desired $y$…I.e the ‘$x$’ that always exists such that $f(x) = y$ is $x = (y - 1)^{1/2}$
Then: $f(x) = f\left((y - 1)^{1/2}\right) = y$
[Edit: Has been kindly noted that this isn't even a valid surjection over the real numbers (which I feel like kind of speaks to my question)...]
$x^3 + 1$ should work...
Question this error speaks to another problem I have. Is there any systematic way of knowing when that 'taking the inverse' method will fail...apart from just spotting an error where the domain and range are not equal?
Question Can anybody suggest some maps where I might think oh this is ‘just self evident’ but actually I need to do some work?
Question In a way I find the ‘self evident’ ones harder, because you kind of don’t need to do anything, but you do need to do something…can anyone give me an exact structure of what I need to say and why it works, and as suggested before, what things not to say (that might seem right) but wouldn’t get credit, or are simply wrong!
Thanks for helping with all the questions, hope this isn’t too basic!
 A: This is how you prove surjectivity (at least, this is probably the most accessible way):

Let $y$ be an arbitrary element of the codomain of $f$. Prove that whatever $y$ is, there exists $x$ in the domain of $f$ such that $y = f(x)$.

In your "self-evident case", $y$ is by definition of the form $a x$ for some $x \in H$.
That is, whatever $y$ is, there is by the definition of a coset some $x$ in $H$ such that $y = f(x)$.
The "self-evidentness" of this is because it follows directly from the definition.
Hence to present such a proof, all you need to do is show "from the definition of (whatever) it follows that $\forall y \in \operatorname{Cdm} (f): \exists x \in \operatorname{Dom} (f): y = f(x)$.
Hope this helps.
Your second example is NOT as it stands a surjection because the image set of $f$ is all real numbers greater than or equal to $1$. It is assumed, unless we are told to the contrary, that if $f$ is defined as a "real function", the codomain is taken to be the whole of the set of real numbers.
If of course you are in the complex plane, then $f$ is of course surjective -- but obviously not injective.
A: Surjectivity can be defined in the following way:

Let $X,Y$ be sets, and let $f:X\to Y$ be a function. We say that $f$ is surjective if $Y=f(X)$.

What this means is that $Y$, also called the codomain, is equal to the range (or image) of $f$. Now by definition we must have that $f(X)\subseteq Y$, and so to prove surjectivity you prove that $Y\subseteq f(X)$. To do this, prove that if $y\in Y$, then $y\in f(X)$, which by definition means that $y=f(x)$ for some $x\in X$. Now some times this is a really trivial matter, such as in your first example, when you already are given that $Y=f(X)$, and in other cases this is extremely non-trivial and requires hard work.
A: I would say that a function $f$ is self-evidently a surjection, when codomain is defined to be the range of $f$
For instance in your left coset example.
We define $aH := \{ah, h\in H\}$ and $F : H\to aH, h\mapsto ah$.
In other words if we denote $f_a : G\to G, g\mapsto ag$,
then we have just said : take $aH = f_a(H)$ and take $F : H\to aH$ to be the restriction of $f_A$ from $H$ to $aH$.
Then by definition, because $aH$ is the image of $H$ by $f_A$, $F$ is surjective.
If you are not in this specific instance, you need a proof.
