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I have an idea of where to go with this proof, but would like a second opinion as to wether I have actually made a logical argument.

$f(g(x)) \neq f(g(y))$ where $x, y \in \mathbb{R}$ and $x \neq y$.

Given that $g(x)$ is a $1$-$1$ function $g(x) \neq g(y)$. For simplicities sake we will say that $g(x) = a$ and $g(y) = b$, where $a, b \in \mathbb{R}$ and $a \neq b$

This leaves us with $f(a) \neq f(b)$. As we have established that $a \neq b$, and we know that $f(x)$ is a $1$-$1$ function this statement is true, hence proving that $f(g(x))$ is a $1$-$1$ function

I also have to establish whether it is true if $f(g(x))$ is $1$−$1$ must $f(x)$ and $g(x)$ be $1$−$1$ on their domains?

To me this is true as the $g(x)$ must be $1$-$1$ for $f(g(x))$ to be $1$-$1$, and if say $f(g(x)) = f(g(-x))$ (i.e. $f(x)$ is not $1$-$1$) then that would contradict the fact that $f(g(x))$ is $1$-$1$.

Any help is greatly appreciated!

Zhanxiong
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Cameron
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    Suppose the domain of $f$ and the range of $g$ are each one-point sets, while the range of $f$ and domain of $g$ is a two-point set. What can you say about $f\circ g$, $f$, and $g$? – rogerl Aug 22 '15 at 13:46
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    why not just say for x and y given, $f(g(x)) = f(g(y)) \implies g(x) = g(y) \implies x = y$? – George Aug 22 '15 at 13:47
  • If $f(g(x))$ is 1-1, then $f$ has to be 1-1. $g$ only has to be 1-1 when restricted to whatever $f$ hits. What happens outside there you cannot say anything about. – Arthur Aug 22 '15 at 13:48
  • I believe in this example the OP is restricting the question to real valued functions of the real line @rogerl – Joel Aug 22 '15 at 13:50
  • Welcome student of University of Adelaide..... – M.S.E Aug 22 '15 at 13:54
  • @Joel OK, then, let $f:\mathbb{R}\to\mathbb{R}:x\mapsto e^x$, and let $g:\mathbb{R}\to\mathbb{R}:x\mapsto x$ for $x\ge 0$ and $g(x)=0$ for $x<0$. Then $g\circ f$ is 1-1 but $g$ is not. Same example as above, different domains etc. – rogerl Aug 22 '15 at 13:57
  • @George: Yea I guess that would probably be easier :P – Cameron Aug 22 '15 at 14:04
  • Your idea is good. But your proof is backwards. You prove $f(a) = f(b) \Rightarrow a = b \Leftrightarrow a \neq b \Rightarrow f(a) \neq f(b)$ to prove injectivity. – Race Bannon Aug 22 '15 at 14:13

1 Answers1

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Let $f : B \rightarrow C$ be bijective.

Let $g : A \rightarrow B$ be bijective.

Injectivity:

$$f(g(x)) = f(g(y)) \Rightarrow g(x) = g(y) \Rightarrow x = y$$

Surjectivity:

$$f(g(A)) = f(B) = C$$

Myridium
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