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Consider two functions $f\colon A \to B$ and $g\colon B \to C$. How can I prove the following?

  • If $f$ and $g$ are one-to-one, then the composition function $g \circ f$ is one-to-one.
  • If $f$ and $g$ are onto functions, then $g \circ f$ is an onto function.
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  • $\begingroup$ What are your thoughts? Do you know what one-to-one or onto means? $\endgroup$ – anon May 6 '13 at 6:45
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For injectivity of $g\circ f$ you have to check that $$g(f(x))=g(f(y))\rightarrow x=y$$ holds for every $x,y\in A$. Use first that $g$ is one-to-one, then that $f$ is one to one.

For surjectivity, given $z$ arbitrary in $C$ you are requested to find an element $x\in A$ such that $g(f(x))=z$. Take a pre-image $y\in B$ of $z$ via $g$, using surjectivity of $g$ and then take a preimage $x\in A$ of $y$ using surjectivity of $f$. Check that $g(f(x))=z$

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