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?
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.
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.