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Question 3.26

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TZ
leumasicOfficial

3 months ago

We have that

h=hI=h(fg)=(hf)g=Ig=g.h=h\circ I=h\circ(f\circ g)=(h\circ f)\circ g=I\circ g=g.
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Q 3.26

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Q 3.26