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Exercise:
Let f:X rightarrow Y g:Y rightarrow Z be maps. The following properties are to be shown: abcliste abc If fg are injective then g circ f is also injective. abc If fg are surjective then g circ f is also surjective. abc If fg are bijective then g circ f is also bijective. abcliste

Solution:
Proof. abcliste abc We need to prove that forall x_ x_ in X the following holds: if gcirc fx_gcirc fx_ then x_x_. So let x_x_ in X be s.t. gcirc fx_gcirc fx_. This is equal to gfx_gfx_. Since g is injective it follows that fx_fx_. Since g is injective as well we can conclude that x_x_. abc We need to show that forall z in Z exists x in X: gcirc fxz. Indeed let z in Z. Since g is surjective exists y in Y: gyz. As f is also surjective it follows that exists x in X: fxy. This allows the conclusion: gcirc fx gfxgyz. abc Follows from a+b because glqq bijective injective+surjectivegrqq abcliste
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Exercise:
Let f:X rightarrow Y g:Y rightarrow Z be maps. The following properties are to be shown: abcliste abc If fg are injective then g circ f is also injective. abc If fg are surjective then g circ f is also surjective. abc If fg are bijective then g circ f is also bijective. abcliste

Solution:
Proof. abcliste abc We need to prove that forall x_ x_ in X the following holds: if gcirc fx_gcirc fx_ then x_x_. So let x_x_ in X be s.t. gcirc fx_gcirc fx_. This is equal to gfx_gfx_. Since g is injective it follows that fx_fx_. Since g is injective as well we can conclude that x_x_. abc We need to show that forall z in Z exists x in X: gcirc fxz. Indeed let z in Z. Since g is surjective exists y in Y: gyz. As f is also surjective it follows that exists x in X: fxy. This allows the conclusion: gcirc fx gfxgyz. abc Follows from a+b because glqq bijective injective+surjectivegrqq abcliste
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Tags
eth, hs22, injective, lineare algebra, proof
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(2, default)
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Language
ENG (English)
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