Is the following a valid mathematical statement? For all $f:\mathbb N\to\{1,2,3,\ldots,100\}$,
 If $f$ is a one to one correspondence,
 Then $f^{-1}(2)=3$
It seems as though this should not be a valid statement, since the implication fails to remain a statement when the hypothesis becomes false. This is because when $f$ is not one to one and onto, the syntax $f^{-1}(2)$ is undefined. So, the implication on a whole becomes undefined under such a condition. Is this true?
 A: The question is more complicated then a simple ex falso quodlibet. If the sequence of letters makes sense at all, then it is true, since there is no one-to-one function $\mathbb{N}\rightarrow\{1, \ldots, 100\}$. 
However, it is not clear whether we are talking about a logical formula at all. For example $(\exists x: x\neq x)\Rightarrow (x+\forall)$ is not a valid formula, in particular it is not true, although it is of the form "false $\Rightarrow$ something".
The real problem is that $f^{-1}$ is a short-hand notation which goes beyond the realm of formal logic. We can declare that $f^{-1}(a)=b$ means that an inverse function exists and at $a$ this inverse function takes the value $b$. If we do so, then the statement becomes true. On the other hand we can view $^{-1}$ as a partial operation on the set of functions, i.e. the inverse function, if such a function exists, and gloppidyglopp otherwise. Then any statement involving $f^{-1}$ for a function $f$, which is not a bijection becomes neither true nor false, but undefined. 
In this case this looks like (and probably is) nitpicking. However, there are real world examples where such a distinction does matter: What is the meaning of the statement $\lim a_n\neq 0$, if the limit does not exist? True, false, and undefined are all valid options.
The moral of the story is probably that when writing mathematics you should try to avoid ambiguities, and when reading mathematics, you should try to get the intended meaning, even if it differs from the formal statement.
A: An instructor asked the class whether all cell phones in the classroom are shut off.
If by some fluke, there are no cell phones in the classroom, then the answer is "yes".  "Yes" means there are no turned-on cell phones in the classroom.
Same thing here: there are no one-to-one correspondences from $\mathbb N$ into a set of only $100$ members; therefore, any statement you could make about such a correspondence is true.
A: "It seems as though this should not be a valid statement"
If it's not a valid statement, then either "f is a one to one correspondence" is true and "f−1(2)=3" is false, or both statements are false and "if p, then p" is false, where the truth value of the proposition "p" is false.  It is not the case that ""f is a one to one correspondence" is true and "f−1(2)=3" is false" holds.  It is also not the case that "if p, then p" is false, when the truth value of "p" is false.  If it were, then "if p, then p" would not qualify as a tautology.  Consequently, the statement is valid.
A: Here are two ways to think about your question going back to the philosophy of logical reasoning:
(1) If we trust that classical propositional-logic is an adequate account of truth-preservation in mathematical reasoning then, looking at the truth-table for the material-conditional in classical propositional-logic, we see that the material-conditional is only false when the antecedent is true and the consequent is false. $Every$ of the 3 remaining cases is going to make the material conditional true. In this case, we have a false antecedent so it doesn't matter what the truth-value of the consequent is - that is, it doesn't matter whether $f^{-1}(2) = 3$ or not since the antecedent is already false. This is the idea behind Michael Hardy's comment.
(2) If (1) is counter-intuitive, as it may very well be as it is counter-intuitive for many people, then consider that $p \rightarrow q$ is logically-equivalent classically-speaking to $\sim p \vee q$. Seen this way, either it is false that there is a bijection from $\mathbb{N}$ to $\{1,2,3,...,100\}$ or it is true that $f^{-1}(2) = 3$. In our case, it is false that there is a bijection from $\mathbb{N}$ to $\{1,2,3,...,100\}$, making $\sim p$ true, so the statement is valid.
