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Suppose $f$ and $g$ are functions that fail to be one-to-one, but $f+g$ is one-to-one. Has anyone ever seen the notation $(f+g)^{-1}$ for the inverse function in that situation? (I find myself wondering if I'm comfortable with this.)

(Example: Consider the six familiar trigonometric function as mappings from $\mathbb R \bmod 2\pi$ to $\mathbb R\cup\{\infty\}$, where the codomain is just the one-point compactification, so that instead of $+\infty$ and $-\infty$ we have one $\infty$ at both ends of the line. All six fail to be one-to-one. But $\sec+\tan$ and $\sec-\tan$ are one-to-one (and onto if we remove the one removable discontinuity from each).)

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    Would try to avoid the notation.2012-12-23
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    So far we have this comment saying one person would try to avoid that notation, and one posted answer saying another would have no hesitation about using it. What would be the pros and cons?2012-12-23
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    Pros: brevity; Cons: brevity.2012-12-23
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    Since I mentioned $\sec+\tan$ and $\sec-\tan$, maybe I should mention that they are each others reciprocals (not each others inverse functions).2012-12-23
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    I don't see anything wrong with it; there's little possibility for confusion. As you note below, nobody would bat an eye if $f$ and $g$ were singular matrices.2012-12-23

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If $h$ is an invertible function, then $h^{-1}$ can be used to denote its inverse. In particular, we can take $h=f+g$ here. It is not relevant if $f$ or $g$ are invertible. I would not hesitate to use this notation myself, though I can't be sure if I have come across it.

In a similar way, we write $\frac{1}{1+0}=1$, even though $\frac{1}{0}$ is undefined in the real numbers!

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    But do we ever write $\dfrac{1}{a+b}$ where $a$ and $b$ both fail to have reciprocals? I imagine that probably that does happen with some sort of objects in the role of $a$ and $b$, although I wouldn't do it with matrices. Certainly with matrices $a$ and $b$ I would have no hesitation about writing $(a+b)^{-1}$; that happens all the time.2012-12-23
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    @MichaelHardy I don't understand your point. You can certainly have two singular matrices $a$ and $b$ such that $a+b$ is nonsingular, and then you can write $(a+b)^{-1}$.2012-12-23
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    @RobertIsrael : That's what I said. Obviously. Are you suggesting that you _would_ write $\dfrac{1}{a+b}$ where $a$ and $b$ are matrices? I've never seen it done and I wouldn't do it.2012-12-23