In Chapter 12 of Spivak's Calculus, the definition of the inverse of function $f$ is provided as:
For any function $f$, the inverse of $f$, denoted by $f^{-1}$, is the set of all pairs $(a,b)$ for which the pair $(b,a)$ is in $f$
Given the above definition, is it fair for me to say that if the symbol '$f^{-1}$' shows up as an assumption, I can automatically assert that $f$ is a function?
The context for this question comes from a Spivak-provided theorem:
If $f^{-1}$ is a function, then $f$ is one-to-one.
Spivak's proof reads:
Suppose that $f^{-1}$ is a function. $\color{red}{\text{If } f(b)=f(c)}$, then $f$ contains the pairs $(b,f(b))$ and $(c,f(c))=(c,f(b))$, so $(f(b),b)$ and $(f(b),c)$ are in $f^{-1}$. Since $f^{-1}$ is a function, this implies that $b=c$. Thus $f$ is one-to-one.
I have highlighted in $\color{red}{\text{red}}$ where my confusion lays. As far as I know, the syntax '$f(x)$' is only to be used if we know that $f$ is a function...so there must be an implicit understanding that $f$ is a function.
Is that correct?
The answer is right in the first clause in the definition: "For any function $f$..."
That means he is only defining $f^{-1}$ on functions. The definition he then gives yields a relations that is itself a function if and only if the original function was injective.
Note: There's no reason you can't define the inverse of a generic relation in the same way and the same definition would work even if $f$ wasn't a function. But since he includes that clause, he's restricting the definition of the symbol $f^{-1}$ to cases where $f$ is a function