$\mathrm{\Psi}$ is surjective if and only if ${\mathrm{\Psi}}^{\ast}$ is injective
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
$\mu f+\lambda g:x$  $\mapsto $  $\mu f(x)+\lambda g(x).$ 
Next, let $Y$ be another set, let $\mathrm{\Psi}:X\to Y$ is a mapping, and let ${\mathrm{\Psi}}^{\ast}:M(Y,V)\to M(X,V)$ be the pullback of $\mathrm{\Psi}$ as defined in this (http://planetmath.org/Pullback2) entry.
Proposition 1.
$$

1.
${\mathrm{\Psi}}^{\ast}$ is linear.

2.
If $V$ is not the zero vector space, then $\mathrm{\Psi}$ is surjective^{} if and only if ${\mathrm{\Psi}}^{\ast}$ is injective^{}.
Proof.
First, suppose $f,g\in M(Y,V)$, $\mu ,\lambda \in F$, and $x\in X$. Then
${\mathrm{\Psi}}^{\ast}(\mu f+\lambda g)(x)$  $=$  $(\mu f+\lambda g)(\mathrm{\Psi}(x))$  
$=$  $\mu f\circ \mathrm{\Psi}(x)+\lambda g\circ \mathrm{\Psi}(x)$  
$=$  $\left(\mu {\mathrm{\Psi}}^{\ast}(f)+\lambda {\mathrm{\Psi}}^{\ast}(g)\right)(x),$ 
so ${\mathrm{\Psi}}^{\ast}(\mu f+\lambda g)=\mu {\mathrm{\Psi}}^{\ast}(f)+\lambda {\mathrm{\Psi}}^{\ast}(g)$, and ${\mathrm{\Psi}}^{\ast}$ is linear. For the second claim, suppose $\mathrm{\Psi}$ is surjective, $f\in M(Y,V)$, and ${\mathrm{\Psi}}^{\ast}(f)=0$. If $y\in Y$, then for some $x\in X$, we have $\mathrm{\Psi}(x)=y$, and $f(y)=f\circ \mathrm{\Psi}(x)={\mathrm{\Psi}}^{\ast}(f)(x)=0$, so $f=0$. Hence, the kernel of ${\mathrm{\Psi}}^{\ast}$ is zero, and ${\mathrm{\Psi}}^{\ast}$ is an injection. On the other hand, suppose ${\mathrm{\Psi}}^{\ast}$ is a injection, and $\mathrm{\Psi}$ is not a surjection. Then for some ${y}^{\prime}\in Y$, we have ${y}^{\prime}\notin \mathrm{\Psi}(X)$. Also, as $V$ is not the zero vector space, we can find a nonzero vector $v\in V$, and define $f\in M(Y,V)$ as
$$f(y)=\{\begin{array}{cc}v,\hfill & \text{if}y={y}^{\prime},\hfill \\ 0,\hfill & \text{if}y\ne {y}^{\prime},y\in Y.\hfill \end{array}$$ 
Now $f\circ \mathrm{\Psi}(x)=0$ for all $x\in X$, so ${\mathrm{\Psi}}^{\ast}f=0$, but $f\ne 0$. ∎
Title  $\mathrm{\Psi}$ is surjective if and only if ${\mathrm{\Psi}}^{\ast}$ is injective 

Canonical name  PsiIsSurjectiveIfAndOnlyIfPsiastIsInjective 
Date of creation  20130322 14:36:03 
Last modified on  20130322 14:36:03 
Owner  matte (1858) 
Last modified by  matte (1858) 
Numerical id  6 
Author  matte (1858) 
Entry type  Theorem 
Classification  msc 0300 