What is the relation between two invertible functions

Question

Lets say that if f(x) and g(x) are invertible.
1- is (f(x)+g(x)) also invertible? 2- is f(g(x)) invertible too?

for the first one lets say that f(x)=x and g(x)=-x then f(x)+g(x)=x+(-x)=0 and f(x)=0 is not invertible. so my conculusion is that f(x)+g(x) can be uninvertible if f(x) and g(x) are invertible. Am I right?

For the second one I have no clue.

Answer 1

I have no clue whether the following answers have made headway. As a result, I'll take a stab at this:

For (1), you had to provide an adequate counterexample.

Now, for (2), consider the following: Let $f$ be a function from set $X$ to set $Y$. Likewise, let $g$ be a function from set $Y$ to set $Z$. Thus, $g\circ f$ is a function from set $X$ to set $Z$. We see that $g\circ f$ has an inverse since we can perform $f^{-1}\circ g^{-1}$ to get back to $X$.

Let's decode that. Does it make more sense to think that $f$ is a bridge from place $X$ to place $Y$ and that $g$ is a bridge from place $Y$ to place $Z$? Well, if we have bridges back---that is, $g^{-1}$ being a bridge from place $Z$ to place $Y$ and $f^{-1}$ being a bridge from place $Y$ to place $X$---is it not self evident that to get from place $Z$ to place $X$ we simply take bridge $g^{-1}$ and then bridge $f^{-1}$, and we call this combined route $f^{-1}\circ g^{-1}$? :)

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For sake of completeness, here's what I just said in the dry-Algebra language:

Let $f: X\to Y$ and $g: Y \to Z$. Thus, $g\circ f: X \to Z$.

If we have $f^{-1}: Y\to X$ and $g^{-1}: Z \to Y$, we have that $f^{-1}\circ g^{-1}: Z \to X$.

Letting the identity map be $I$, we have $f^{-1}\circ g^{-1}\circ g\circ f=I$. Therefore, $f^{-1}\circ g^{-1}$ is the inverse map of $g \circ f$.

Answer 2

HINT: You are right about (1). For (2), suppose that you know that $f\big(g(x)\big)$ is some particular number $y$. Does the invertibility of $f$ guarantee that in principle you can find $g(x)$? And what then does the invertibility of $g$ tell you?

Answer 3

You are right about 1. 2 is correct provided $Range(g) \subseteq dom(f)$. In any case it has to be or else $f(g(x))$ would not be well defined.