I've already proven that if we assume f is bijective and g is bijective, then $(g \circ f)$ is bijective. I've also proven that$(g \circ f)^{-1}$ exists. I'm stuck on this part, however. Any suggestions?
Prove that $(g \circ f )^{-1} = f^{-1} \circ g^{-1}$
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elementary-set-theory
proof-writing
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0Do you know the rule for groups? The proof here is the same. – 2012-12-04