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Suppose we have $V$ and $W$ be $2$ $K$-vectorspaces and $f:V\rightarrow W$ is a linear map then:

  1. For all $\varphi\in W^\ast$ one has $\varphi\circ f\in V^*$
  2. The map $f^\ast:W^\ast\rightarrow V^\ast:\varphi\mapsto\varphi\circ f\;$ is a linear map, which is also called a dual map.

Where $V^\ast$, the dual space, is defined as: \begin{equation} V^\ast:= Hom_K(V,K) = \{\;f:V\rightarrow K\;|\;f\;\text{is a linear map}\;\} \end{equation}

Where $K$ is a $1$-dimensional $K$-vector space and $V$ a $K$-vector space

But I don't really get what this means, could anyone give a clear explanation because I have a hard time understanding this. Especially with that extra $\varphi$ function

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The set $V^*$ of all linear maps from a vector space $V$ to its base field $K$ is, itself, a vector space over $K$: this is easy to show, with the operations being $(f+g)(x) = f(x) + g(x)$ and $(\lambda f)(x) = \lambda(f(x))$, for all $f, g\in V^*$ and all $\lambda \in K$.

Now, given any $\varphi \in W^*$, and any $f: V \to W$, $\varphi\circ f$ is a linear map from $V$ to $K$ (since it's a composition of linear maps), so is an element of $V^*$ (you have a typo in your question: it is not an element of $V$).

Thus, we define a new function, $f^*$, that takes as its input an element $\varphi$ of $W^*$, and spits out an element of $V^*$, by composing $\varphi$ with $f$. This, it turns out, is linear, and this, too, is not hard to check: if $\varphi,\psi\in W^*$, and $v \in V$, then

\begin{align*}f^*(\varphi+\psi)(v) &= (\varphi + \psi)\circ f(v) \\&= (\varphi+\psi)(f(v)) \\&= \varphi(f(v)) + \psi(f(v)) \\&= \varphi\circ f(v) + \psi \circ f(v) \\&= (f^*\varphi + f^*\psi)(v),\end{align*}

hence $f^*(\varphi + \psi) = f^*\varphi + f^*\psi$. Similarly, for any $\lambda \in K$, we have

\begin{align*}f^*(\lambda\varphi)(v) &= (\lambda\varphi)\circ f(v) \\&= (\lambda\varphi)(f(v))\\&=\lambda(\varphi(f(v))) \\&= \lambda(\varphi \circ f)(v)) = \lambda f^*(\varphi)(v),\end{align*}

hence $f^*(\lambda\varphi) = \lambda f^*(\varphi)$, and $f^*$ is, indeed, linear.

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