Definition of one-to-one functions and inverse I was wondering if anyone knows a book that defines one-to-one similar to the below definition? This is taken from my lecture notes but I need a book that has this definition because this definition seems a bit strange and hard for me to understand so I need a book that explains this well. The definition that I used to know was that if $f(a)=f(b)$ implies that $a=b$, then $f$ is one-to-one and $f$ has an inverse if $f$ is one-to-one.

Let $f:A\rightarrow B$ be a function with the domain $A$ and codomain
  $B$. Let $y\in B$, and solve $$ f(x)=y $$ for $x\in A$. There are three
  possibilities:
  
  
*
  
*No solution for $x\in A$ ($y$ is not in the range of $f$)
  
*Exactly one solution for $x\in A$ ($y$ is in the range of $f$)
  
*More than one solution for $x\in A$ ($y$ is in the range of $f$)
  
  
  If either case 1 or 2 happens, we call $f$ one-to-one and if only case
  2 happens, we say that $f$ has an inverse.

 A: I'm not aware of a book using this particular presentation of the definition, but this definition is equivalent to the one you are used to. 
So, essentially we need to make sure that the property of every $y\in B$ satisfying either condition $(1)$ or $(2)$ is equivalent to injectivity. Suppose $y$ satisfies $(1)$. Then $f^{-1}(y)=\{x\in A:f(x)=y\}=\emptyset.$ Now, it is true that $f(a)=y=f(a')$ implies $a=a'$ because there do not exist such $a,a'\in A$ satisfying the equation.  This is a case of vacuous logic, which might seem a little confusing. The takeaway is that there is no counterexample to this "for all" statement, so it is true.
Now, suppose that $y$ satisfies $(2)$, then $f(a)=f(a')=y$ implies $a=a'$, or else there are two solutions to $f(x)=y$. This agrees with the definition you are used to. So, if $y\in B$ implies $y$ satisfies either of these properties for all $y$, this function is injective in the way you describe.
And of course, if some $y\in B$ satisfies $(3)$, then $f^{-1}(y)$ has multiple distinct values- at least $2$. Call some pair of them $a,b$. Then from $a\ne b$, we have $f(a)=f(b)=y$ does not imply $a=b$. This contradicts injectivity by your definition.
I'll leave it up to you to show that the converse is true, i.e. that your definition implies this one. Hopefully this makes some sense: let me know if this is confusing.
