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Let $f: A \rightarrow B$. Suppose $g, h:B \rightarrow A$ so that $f \circ g = I_B$ and $h \circ f = I_A$. Show that $f$ is a bijections and $g=h=f^{-1}$.

$I_A $ and $ I_B$ denote the identity functions for sets $A$ and $B$.

I've been working on this one for a while now, and don't really understand how to show it

tzamboiv
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2 Answers2

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$$h=h\circ I_B=h\circ(f\circ g)=(h\circ f)\circ g=I_A\circ g=g$$

surjectivity:

$b=f(g(b))$ for each $b\in B$

injectivity:

If $f(a_1)=f(a_2)$ then $a_1=h(f(a_1))=h(f(a_2))=a_2$

drhab
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  • After the first line of this answer, it would suffice to say: the map $g=h$ satisfies the defintion of inverse map of $f$, and by its existence $f$ is automatically bijective. One does not even have to mention elements (the proof is valid in any category, if one replaces "bijection" by "isomorphism"). – Marc van Leeuwen Oct 09 '15 at 08:49
  • @Marc That is correct. It is informative though in the sense that it shows on what this "automatically bijective" is based. Actually it shows that in category Sets the isomorphisms are exactly the bijections. It might be redundant, but that also depends on the knowledge of the OP. – drhab Oct 09 '15 at 10:22
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Hint:

To show that $f$ is a bijection, it suffices to show that it's injective and surjective. Try to use $f\circ g = I_B$ to show that $f$ is surjective and $h\circ f = I_A$ to show that $f$ is injective.