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What would be an isomorphism between $\mathcal{F}(S;V / U)$ and $\mathcal{F}(S;V)/ \mathcal{F}(S;U)$, where $S$ is a set, $V$ a vector space and $U$ a subspace of $V$. $\mathcal{F}(A,B)$ denotes the set of functions from $A$ to $B$ and $V/U$ is the quotient space.

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To produce such an isomorphism we have to get a hold on the elements of $V/U$ without explicitly talking about a complement of $U$ in $V$.

A general element of $V/U$ is represented by elements $v\in V$, where $v$ and $v'\in V$ represent the same element $p\in V/U$ iff there is an $u\in U$ such that $v'=v+u$.

Therefore a general element $\Phi\in {\cal F}(S,V/U)$ is represented by functions $\phi:\ s\mapsto v(s)\in V$, and any $\phi\in{\cal F}(S,V)$ can be used to represent such a $\Phi$. Two functions $\phi$, $\phi'\in{\cal F}(S,V)$ represent the same $\Phi:\ S\to {\cal F}(S,V/U)$ iff for each $s\in S$ the values $\phi(s)$ and $\phi'(s)$ are equivalent mod $U$, i.e., iff for each $s\in S$ there is an $u=:u(s)\in U$ such that $\phi'(s)=\phi(s)+u(s)$. This is saying that the two functions $\phi$, $\phi'\in{\cal F}(S,V)$ differ by a $\theta\in {\cal F}(S,U)$.

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A $\varphi\in \mathcal F(S;V)$ can be viewed as a sequence indiced by the elements of $S$, and accordingly write $\varphi(s)$ as the $s^{\text{th}}$ coordinate $\varphi_s$. Apply the canonical $V\to V/U$ 'coordinatewise', giving an $\mathcal F(S;V)\to\mathcal F(S;V/U)$, see that its surjective and find its kernel.

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    I see what you mean. The kernel will be $\mathcal{F}(S;U)$.2012-09-29