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I have seen questions on *.stackexchange.com that pertain to finding the inverse of functions that map from $\mathbb{R}^2$ to $\mathbb{R}^2$ (for example, this, and this).

My question pertains to finding the inverse of a function like the perspective function, defined as: $P:\mathbb{R}^{n+1}\rightarrow\mathbb{R}^n,\, P(x,\,t) = \frac{x}{t},\, \text{dom } P = \{(x,\, t)\,\vert\,(x \in \mathbb{R}^n)\wedge(t>0)\}$.

Thus, my question is: how can one find the inverse of this function? I am interested in doing so because I have recently learned that the inverse-image of a convex set under this function is also convex.

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You cannot find inverses of functions which are not injective. Your function is not injective, since

$$P(x, t) = P(\alpha x, \alpha t)$$ for any $\alpha > 0$. For example, $$P(2x, 2t) = \frac{2x}{2t} = \frac xt = P(x,t)$$

and even more concretely, for $n=1$, you have $$P(2,1) = P(4,2) = P(6,3)\cdots = 2$$ so it is impossible to define the inverse of this function, since $P^{-1}(2)$ would have to be equal to $(2,1), (4,2), (6,3)$ and many other pairs at the same time.

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    $\begingroup$ I forgot about this concept: it is intuitive, if there are many possible "outputs" for a given "input" in the range of the function, things aren't well-defined. Thanks for the insight, I will accept your answer after the time threshold has passed. $\endgroup$ – Nurmister Sep 17 '18 at 12:36
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    $\begingroup$ One question, then (if it is appropriate to ask here): given the perspective function is not injective, is it meaningless to talk about the inverse image of a set under this function, since it cannot exist? $\endgroup$ – Nurmister Sep 17 '18 at 12:38
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    $\begingroup$ @Nurmister Thanks! Just an additional side note: there do exists injective functions mapping from $D\subseteq \mathbb R^{n+1}$ to $\mathbb R$. Your particular one is not injective, but some are. $\endgroup$ – 5xum Sep 17 '18 at 12:38
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    $\begingroup$ @Nurmister The inverse image of a set is always defined, and it is always a set. By definition, $P^{-1}(A) = \{(x,t)\in \text{dom}(P)| P(x,t) \in A\}$. Note that a function $f$ is invertible if and only if the inverse image of every singleton in the range of $f$ is also a singleton. $\endgroup$ – 5xum Sep 17 '18 at 12:40

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