$\mathrm{\Psi }$ 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 withpointwise 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 aninjection.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 canfind a non-zero vector $v\in V$, and define $f\in M(Y,V)$ as

Now $f\circ \mathrm{\Psi }(x)=0$ for all $x\in X$, so ${\mathrm{\Psi }}^{\ast }f=0$,but $f\ne 0$. ∎