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Consider these two diagrams of sets and functions (with $f$ and $f'$ invertible):

\begin{array}{ccccccccc} A & \overset{f}{\longrightarrow} & B && &B & \overset{f^{-1}}{\longrightarrow} & A\\ u\downarrow& & v\downarrow & &; & v\downarrow & & u\downarrow\\ A' & \overset{f'}{\longrightarrow} & B' & && B' & \overset{f'^{-1}}{\longrightarrow} & A' \end{array}

Can I say that the first commutes if and only if the second does? In other words, is it true that

$$v\circ f=f'\circ u \textrm{ if and only if } u\circ f^{-1}=f'^{-1}\circ v$$

Danae Kissinger
  • 2,143

1 Answers1

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Sure. $v\circ f =f' \circ u\Rightarrow f'^{-1}\circ v= f'^{-1}\circ v \circ f\circ f^{-1}=f'^{-1}\circ f'\circ u \circ f^{-1}=u\circ f^{-1}$

alpacahaircut
  • 367