Inflation $I:(V/U)^* \to V^*$

Question

For a (finite) vector space $V$ and subspace $U \subset V$, my exercise sheet defines a function called an Inflation with $$I: (V/U)^* \to V^*$$ such that $$(Ig)(v) = g(v + U) ~~~~~~\forall v\in V ~~\text{and}~~g\in(V/U)^*.$$

I am confused why $g(v + U) \in V$? We want $I(g) \in V^*$ which means $I(g): V \to V$ but $$g(v + U) \notin V$$ but rather $$g(v + U) \in (V/U).$$ since $g: (V/U) \to (V/U)$.
This seems contradictory. What am I missing?

Answer 1

We want $I(g) \in V^*$ which means $I(g): V \to V$

No, the definition of $V^*$ is the space of linear maps from $V$ to the underlying field: $V\to \mathbb F$.

Answer 2

The map $I$ you want to define has domain $(V/U)^*$ and codomain $V^*$.

You want to define its action on $g\in (V/U)^*$ (so $g$ is a linear form on $V/U$): $Ig$ should be a linear form on $V$ and therefore you need to define its action on every element of $V$. The natural choice is $$ Ig\colon v\mapsto g(v+U) $$ because you know that $g(v+U)\in F$ (where $F$ is the field of scalars).

Now you have to prove:

1. for every $g\in (V/U)^*$, $Ig\in V^*$ (that is, it is a linear map $Ig\colon V\to F$);

2. $I$ is linear.

Both statements are easy verifications.

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There is a different way to see this. Let $T\colon V\to W$ be a linear map. Then we can define $T^*\colon W^*\to V^*$ by defining, for $g\in W^*$, $$ T^*g\colon v\mapsto g(Tv) $$ In other words, $T^*g=g\circ T$ (map composition), which spares you verifying point 1 above.

The particular case of the inflation is when you consider $T\colon V\to V/U$, the canonical projection.