Isomorphism of Hom?

Question

How do you show that $$\operatorname{Hom} (\mathbb{R}^n, \mathbb{R}) \simeq \mathbb{R}^n$$?

What is the explicit isomorphism?

I'm trying to understand the concept of the cotangent bundle $T^{*}M$ for a manifold $M$, and I'm told that $T^{*}M$ is just the $\mathbb{R}$-vector space $\operatorname{Hom} (\mathbb{R}^n, \mathbb{R})$, so that for each vector in $\mathbb{R}^n$, we have an associated element in $\operatorname{Hom} (\mathbb{R}^n, \mathbb{R})$. I suppose I'm a bit confused as to why this defines the cotangent bundle properly. I know that $T^{*}M$ is the set of pairs $(x,\eta) \in M \times T^{*}M$. How does this isomorphism give our definition of a linear functional in this cotangent bundle?

Answer 1

This is a special case of the usual result

If $V$ is a finite-dimensional vector space over a field $K$ then $V\cong\mathrm{Hom}(V,K)$

How do we get the isomorphism? Pick a basis $v_1,\dots,v_n$ of $V$. Then let $v_i^*:V\to K$ be the map sending $\sum \lambda_j v_j$ to $\lambda_i$.

Then you can show that $v_1^*,\dots,v_n^*$ is a basis for $\textrm{Hom}(V,K)$, and the map $$\sum \lambda_i v_i \mapsto \sum \lambda_i v_i^*$$ is the desired isomorphism.

Importantly, you do need to pick a basis first in order to get an isomorphism - there is no canonical isomorphism between $V$ and $\textrm{Hom}(V,K)$, just one corresponding to each choice of basis on $V$.

Answer 2

An explicit isomorphism $F:\hom(\mathbb R^n,\mathbb R)\to \mathbb R^n$ is given by

$$F(A)=(A(e_1),\dots,A(e_n))$$

where $A\in\hom(\mathbb R^n,\mathbb R)$ and $e_1,\dots,e_n$ is the canonical basis of $\mathbb R^n$.

Its inverse is $$v\to A_v\qquad A_v(w)=\langle v,w\rangle$$ where $\langle\cdot,\cdot\rangle$ denotes the standard scalar product of $\mathbb R^n$