Pre-images, one to one functions and inverse function.

I have a function defined by:

$$f = \{(1, 5), (2, 5), (3, 6), (4, 6), (5, 7)\}$$

I am asked to find the pre-image of $$6$$. Now I know that the definition of pre-image is $$f^{-1}(y) = x$$. Hence, I am looking for $$f^{-1}(6)$$. However, I am wondering if this means that in order for there to be a pre-image you need for the inverse function to in fact be an inverse function. In this case, since $$f$$ is not one to one, I know it cannot possibly have an inverse function.

My question is, for this reason, do we say that there is no pre-image? Or do I report that there are two pre-images ($$4$$ and $$3$$)?

• Please read this tutorial on how to typeset mathematics on this site. – N. F. Taussig Nov 4 '18 at 18:44

In words, the pre-image of $$6$$ is "the set of all things which map to $$6$$. In this case (if I understand your notation correctly) you have two elements which map to $$6$$. These are not two distinct pre-images. The pre-image is a set, and in this case it is a set which contains two elements.
Another question might be "What is the pre-image of $$6.5$$? In this case, nothing maps to $$6.5$$, and so the pre-image is empty, and you would write the empty set as your answer.
You are correct in your assertion that your function does not have a well-defined inverse, since there are two elements which map to $$6$$, but that doesn't stop you from evaluating what the pre-image of $$6$$ is.
You should check the definition of preimage used in your text/course. The preimage $$f^{-1}(y)$$ of $$y \in Y$$ of a function $$f:X\to Y$$ is usually defined as (something equivalent to): $$\left\{x \vert x \in X \wedge f(x)=y\right\}$$ so it is the set of all $$x$$-values being mapped to $$y$$, so you would 'report' it as the set $$\left\{3,4\right\}$$.