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How is there a bijective map between the sets $\mathcal{L}(V,W) = \{T ~|~ T:V \to W\}$ and $W^\mathcal{B} = \{T ~|~ T: \mathcal{B} \to W\}$ where $T$ is a linear function and $\mathcal{B}$ is a basis for $V$.

Denote $\mathcal{B} = \{b_1,b_2, \cdots, b_n\}$. It makes sense that for any $v \in V$, $v = \sum \alpha_i b_i$ and so each $v$ is composed of a unique tuple $(a_1,a_2, \cdots, a_n)$ to ensure the injectivity criteria. For surjectivity, we have the spanning property of the basis that does the work.

But this doesn't make much sense to me because $T(v) \in W$ where $v\in V$ and $T(\sum \alpha_i b_i) \in W$ where $\sum a_i b_i \notin \mathcal{B}$.

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  • $\begingroup$ $\mathcal{L}(V,W)$ is not the set of all functions from $V$ to $W$ but the set of all the linear functions from $V$ to $W$. $\endgroup$ – Robin Carlier Nov 6 '19 at 18:17
  • $\begingroup$ In $\mathcal{L}(V,W)$ you take all linear maps. In $W^\mathcal{B}$ you take all maps. the symbol $\cong$ denotes a bijection. This is what happens when we choose a basis $\mathcal{B}$ for the vector space $V$. $\endgroup$ – GEdgar Nov 6 '19 at 18:19
  • $\begingroup$ So is the notation $\{T ~| ~T : V \to W\} \cong \{T ~|~ T : \mathcal{B} \to W\}$ correct? I'm sorry I still don't get it $\endgroup$ – charlesh Nov 6 '19 at 18:28
  • $\begingroup$ @charlesh No, it is not. $\{T\, |\, T: V \to W\}$ denotes the set of all maps from $V$ to $W$ while $\mathcal{L}(V, W)$ denotes only the set of linear maps. $\endgroup$ – Robin Carlier Nov 6 '19 at 18:32
  • $\begingroup$ I forgot to mention that $T$ is a linear function. Will it satisfy then? That, $\mathcal{L}(V,W) = \{T | T:V \to W\}$ and $W^\mathcal{B}= \{T | T : \mathcal{B} \to W \}$. I have edited the original post as well $\endgroup$ – charlesh Nov 6 '19 at 18:40
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Let's clear the misunderstanding about the notations by writing explicitly the sets, and why they are in bijection.

Say $\mathcal{B}$ is a basis of $V$. Then, the set $W^{\mathcal{B}}$ is the set $\\{T|T: \mathcal{B} \to W\ \text{is a map} \\}$. This is a set of all maps, without restriction, from the finite set $\mathcal{B}$ to $W$.

The set $\mathcal{L}(V, W)$ can be written as $\\{T | T: V \to W, T\ \text{is linear}\\}$.

Now why are those sets in bijection? Let $T$ be a linear map, and $v \in W$, as you noted, $v = \sum a_i b_i$ for some unique $n$-uple $(a_1,\ldots,a_n)$. Since $T$ is linear, $T(v) = \sum a_i T(b_i)$. So the information about the values of $T$ on $b_i$ is enough to reconstruct, by linearity, the value of $T$ for any element of $V$. This is why the sets are in bijection. In fact, both sets are vector spaces and you can even prove they are isomorphic as such.

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