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  1. Suppose that $f: A→B$ and $g:B→C$ are functions such that $g◦f$ is injective. Prove that $f$ must be injective.

  2. Construct a bijective function $f:R→ (R\setminus \{0\})$. Prove that your function is actually a bijective function.

Can someone help me on how do I prove it?

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closed as off-topic by Scientifica, Gibbs, José Carlos Santos, Saad, Xander Henderson Oct 4 '18 at 11:18

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    $\begingroup$ For two, try the function $$f(x)=\begin{cases}x&x\notin\mathbb N\cup\{0\}\\x+1&x\in\mathbb{N}\cup\{0\}\end{cases}$$ $\endgroup$ – Don Thousand Oct 3 '18 at 22:00
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Given a value $f(x) = f(y) \in B$, since $g$ is a function,

$$g(f(x)) = g(f(y).$$

By the definition of injection of $g\circ f$, $x=y$.

So we have proven that $$f(x)=f(y) \implies x = y,$$

i.e. $f$ is injective.

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Suppose f is not invective.

There exists some $x,y$ in $A$ such that $f(x)= f(y)$

It which case $(g\circ f)(x) = (g\circ f)(y)$

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