Suppose $X$ is a set and $V$ is a vector space over a field $F$ . Let us denote by $M(X,V)$ the set of mappings from $X$ to $V$ . Now $M(X,V)$ is again a vector space if we equip it with pointwise multiplication and addition. In detail, if $f,g\in M(X,V)$ and $\mu,\lambda\in F$ , we set \begin{eqnarray*} \mu f+\lambda g\colon x&\mapsto& \mu f(x) + \lambda g(x). \end{eqnarray*} Next, let $Y$ be another set, let $\Psi\colon X\to Y$ is a mapping, and let $\Psi^\ast\colon M(Y,V)\to M(X,V)$ be the pullback of $\Psi$ as defined in this entry.

Proof. First, suppose $f,g\in M(Y,V)$ , $\mu,\lambda \in F$ , and $x\in X$ . Then \begin{eqnarray*} \Psi^\ast(\mu f+\lambda g)(x) &=&(\mu f+\lambda g)(\Psi(x)) \\ &=&\mu f\circ \Psi(x)+\lambda g\circ\Psi(x) \\ &=&\left(\mu \Psi^\ast(f)+\lambda \Psi^\ast(g)\right)(x), \end{eqnarray*}so $\Psi^\ast(\mu f+\lambda g) = \mu \Psi^\ast(f)+\lambda \Psi^\ast(g)$ , and $\Psi^\ast$ is linear. For the second claim, suppose $\Psi$ is surjective, $f\in M(Y,V)$ , and $\Psi^\ast(f)=0$ . If $y\in Y$ , then for some $x\in X$ , we have $\Psi(x)=y$ , and $f(y)=f\circ\Psi(x)=\Psi^\ast(f)(x)=0$ , so $f=0$ . Hence, the kernel of $\Psi^\ast$ is zero, and $\Psi^\ast$ is an injection. On the other hand, suppose $\Psi^\ast$ is a injection, and $\Psi$ is not a surjection. Then for some $y'\in Y$ , we have $y'\notin \Psi(X)$ . Also, as $V$ is not the zero vector space, we can find a non-zero vector $v\in V$ , and define $f\in M(Y,V)$ as

Now $f\circ\Psi(x)=0$ for all $x\in X$ , so $\Psi^\ast f=0$ , but $f\neq 0$ .