Cfrgtkky

I understand the idea of surjectivity and its definition:
"A function f is surjective if f:A->B if $forall yin Y,exists x in X$ such that $y=f(x)$"
However I have a question on the following excerpt from a textbook:
"Given the function f:R->R where $f(x)=x+1$ determine whether the function is a surjection.(R for the set of all real numbers)

Solution: for sujectivity we rewrite the definition as a universal implication:

$yin R implies exists x in R $ such that $y=f(x)$
'
I understand up to this point. But my issue is with the following:

' then we put the definition of the function into this implication: $yin RLeftarrowRightarrow y=f(x)LeftarrowRightarrow y=x+1LeftarrowRightarrow x=y-1$ which tells us that given y in the codomain R then $y=f(x)$ if $x=y-1$. Hence f is surjective'

My question refers to the first part of the implication " $yin R LeftarrowRightarrow y=f(x)LeftarrowRightarrow y=x+1$"
How/ does y being in the codomain imply that y=f(x) ? (does it?)

We are given $f:mathbb Rtomathbb R$ given by $f(x)=x+1$, and we wish to see if it's surjective. As the book says, the meaning of "$f$ is surjective" here could be written $yinmathbb Rimpliesexists xinmathbb Rtext{ s.t. }y=f(x)$ (with an implied $forall y$).

Then the next part of the book is written in a concise way that I agree is a bit confusing. Here is a rewrite with more words that might clarify what they intended. It's way more words than would be required for a proof, but I wanted to make sure I clarify what's going on.

Let's hope this function is surjective, and so let's use the definition to write in what we hope to prove. Substituting in the definition for $f(x)$ (in other words, using the fact that $y=f(x)iff y=x+1$), we see that we want "$yinmathbb Rimpliesexists xinmathbb Rtext{ s.t. }y=x+1$". Note that $y=x+1iff x=y-1$, so it's enough to prove "$yinmathbb Rimpliesexists xinmathbb Rtext{ s.t. }x=y-1$". But now this is kind of obvious, If $y$ is a real number, then $y-1$ is certainly a real number that equals $y-1$.

Since this is true, the equivalences mean that the original goal "$f$ is surjective" is true as well. $square$

The important facts of the discussion above might be summarized $$yinmathbb R iff exists xinmathbb Rtext{ s.t. }y=f(x)iffexists xinmathbb Rtext{ s.t. }y=x+1iffexists xinmathbb Rtext{ s.t. }x=y-1$$ where the first forward implication you were worried about is only known at the end, since the argument actually went something like $$yinmathbb R implies exists xinmathbb Rtext{ s.t. }x=y-1iffexists xinmathbb Rtext{ s.t. }y=x+1iff exists xinmathbb Rtext{ s.t. }y=f(x)$$ (and $exists xinmathbb Rtext{ s.t. }y=f(x)implies yinmathbb R$ is just a side comment following from the fact that $mathbb R$ is the codomain).

To add on to the above confusion about the order of implications, the author left implicit all of the $exists x$ quantifiers in the chain of reasoning.