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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 $3$)?

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    $\begingroup$ Please read this tutorial on how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Nov 4 '18 at 18:44
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No, we do not say that there is no pre-image. You do not need to have an inverse to have a pre-image.

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.

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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)?

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 $\left\{3,4\right\}$.

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