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I think $fg\colon A \to C$ is bijective as well but I am not sure. What do you guys think?

Any help would be really appreciated.


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If $fg$ is a bijection, then $g=f^{-1}fg$ is also a bijection. – wj32 Feb 23 '13 at 21:44
Try breaking it up: Can you prove $fg$ is surjective? Injective? Since $f$ has an inverse, what can you say about the existence of $(fg)^{-1}$? – andybenji Feb 23 '13 at 21:44
i understand why it should only be injective. What do you think will happen if $f$ was reduced to being surjective only? – uh1 Feb 23 '13 at 22:34
up vote 3 down vote accepted


  • There are injective functions that are not bijective.
  • An example of a bijective function $f:B\to C$ is when $B=C$ and $f$ is the identity map.
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Hint The composition of injective functions is always injective. Think what happens when $B=C$ and $A$ has a strictly smaller size.

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