# Inverse function notation

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. – André Nicolas Dec 23 '12 at 1:34
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? – Michael Hardy Dec 23 '12 at 1:38
Pros: brevity; Cons: brevity. – André Nicolas Dec 23 '12 at 1:43
Since I mentioned $\sec+\tan$ and $\sec-\tan$, maybe I should mention that they are each others reciprocals (not each others inverse functions). – Michael Hardy Dec 23 '12 at 2:02
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. – user7530 Dec 23 '12 at 2:28

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!
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. – Michael Hardy Dec 23 '12 at 1:35
@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 $(a+b)^{-1}$. – Robert Israel Dec 23 '12 at 3:56
@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. – Michael Hardy Dec 23 '12 at 4:01