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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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    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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    @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