Showing that these maps are well-defined

Question

Let $E$ be a (possibly infinite-dimensional) vector space and $F\subseteq E$ a closed linear subspace. The maps $\Phi : (E^*/F^\bot)\rightarrow F^*$ given by $\Phi(f + F^\bot) = f|_F$ and $\Psi:(E/F)^*\rightarrow F^\bot$ given by $\Psi(f) = f(\cdot + F)$ are well-defined.

I know what "well-defined" means, but I can't follow the proofs.

• For $\Phi$ the proof is: If $f + F^\bot = g + F^\bot$, then there is $h\in F^\bot$ such that $f = g + h$. So $$\Phi(f + F^\bot) = f|_F = (g + h)|_F = g|_F + h|_F = g|_F = \Phi(g + F^\bot)$$ since $h(x) = 0$ for all $x\in F$. Hence, $\Phi$ is well-defined.
• For $\Psi$ the proof is: $\vert\Psi(f)(x)\vert = \vert f(x + F)\vert \leq \Vert f\Vert \Vert x + F\Vert \leq \Vert f\Vert \Vert x\Vert$, so $\Psi(f)\in E^*$. Now if $x\in F$, then $$\Psi(f)(x) = f(x + F) = f(F) = f(0 + F) = f(0_{E/F}) = 0$$ as $f\in (E/F)^*$. Hence $\Psi$ is well-defined.

I don't see any pattern here. Of course, these are two different functions, but if I was now faced with a third function, I would not know how to start or what to do. I would be very happy if someone could explain me what is going on here so that I know what to do if I was given a third function and had to show it is well-defined.

Answer 1

I think “well-defined” means two different things in the two statements.

1. To say $\Phi$ is well-defined means that, even though the formula uses a representative $f$ of the coset $f + F^\perp$, the answer doesn't change if we choose a different representative.

2. To say $\Psi$ is well-defined means that even though $\Psi(f)$ is defined as an element of $E^*$, it actually belongs to the subspace $F^\perp$.

If you like, the “well-defined-ness” is needed on the domain side of the first and the codomain side of the second.

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There's a more functorial way to prove these statements. Any time $T \colon W \to W$ is a linear transformation of vector spaces, there is a contravariant dual map $T^* \colon W^* \to V^*$ defined by $T(\lambda)(v) = \lambda(Tv)$. We are given a vector space $E$ and subspace $F$, so there are canonical maps given by the inclusion $\iota \colon F \to E$ and quotient $\psi \colon E \to E/F$. Notice that $\iota^*\colon E^* \to F^*$ is the restriction map $f \mapsto f|_F$, so $F^\perp$ is the kernel of $\iota^*$.

Now notice $\Psi = \psi^*$. Since $\psi \mathrel{\circ} \iota = 0$, we have $\iota^* \mathrel{\circ} \psi^* = 0$. This means that restricting $\psi^*(f)$ to $F$ is the zero map, which is precisely what is needed for $\Psi(f)$ to be in $F^\perp$.

Another quotient map is $\phi \colon E^* \to E^*/F^\perp$. Quotient spaces have the universal property that any map from the full space, whose kernel contains the subspace, factors through the quotient. The map $\Phi$ is the factoring map associated to $\iota^*$, by which we mean $\Phi \mathrel{\circ} \phi = \iota^*$.