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https://texercises.com/exercise/quotient-space-and-complements/
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Exercise:
Let V be a vector space over K and Usubseteq V a subspace. Let Wsubseteq V be a complement of U i.e. Wsubseteq V is a subspace Wcap U and W+UV. Under the above asptions exists a canonical isomorphism Q:Wlongrightarrow V/U defined by Qw:win V/U quad forall win W The isomorphism Q is canonical only once W is specified. In general there is no canonical choice of a complement W to U. In general exists many choices of complements W to U and there is no prefered choice.

Solution:
Proof. Consider the inclusion map i:Wlongrightarrow V iw:win Wsubseteq V and the projection map pi: Vlongrightarrow V/U. Recall that piv:v Longrightarrow Qpi circ i Longrightarrow Q is a linear map. We claim that textKerQ. Indeed if win textKerQ then w Longrightarrow win U. But win W Longrightarrow win Wcap U Longrightarrow w. This show textKerQ. We also claim that Q is surjective. Indeed let xin V/U. Choose vin V s.t. xv. Since VW+U exists win W uin U s.t. vw+u Longrightarrow wvx. so xQw. This proves Q is surjective.
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Exercise:
Let V be a vector space over K and Usubseteq V a subspace. Let Wsubseteq V be a complement of U i.e. Wsubseteq V is a subspace Wcap U and W+UV. Under the above asptions exists a canonical isomorphism Q:Wlongrightarrow V/U defined by Qw:win V/U quad forall win W The isomorphism Q is canonical only once W is specified. In general there is no canonical choice of a complement W to U. In general exists many choices of complements W to U and there is no prefered choice.

Solution:
Proof. Consider the inclusion map i:Wlongrightarrow V iw:win Wsubseteq V and the projection map pi: Vlongrightarrow V/U. Recall that piv:v Longrightarrow Qpi circ i Longrightarrow Q is a linear map. We claim that textKerQ. Indeed if win textKerQ then w Longrightarrow win U. But win W Longrightarrow win Wcap U Longrightarrow w. This show textKerQ. We also claim that Q is surjective. Indeed let xin V/U. Choose vin V s.t. xv. Since VW+U exists win W uin U s.t. vw+u Longrightarrow wvx. so xQw. This proves Q is surjective.
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eth, hs22, lineare algebra, proof, quotient space, vector space
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(3, default)
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ENG (English)
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Proof
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