$\Psi$ is surjective if and only if $\Psi^{\ast}$ 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

 $\displaystyle\mu f+\lambda g\colon x$ $\displaystyle\mapsto$ $\displaystyle\mu f(x)+\lambda g(x).$

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 (http://planetmath.org/Pullback2) entry.

Proposition 1.

$\$

1. 1.

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

2. 2.

If $V$ is not the zero vector space, then $\Psi$ is surjective if and only if $\Psi^{\ast}$ is injective.

Proof.

First, suppose $f,g\in M(Y,V)$, $\mu,\lambda\in F$, and $x\in X$. Then

 $\displaystyle\Psi^{\ast}(\mu f+\lambda g)(x)$ $\displaystyle=$ $\displaystyle(\mu f+\lambda g)(\Psi(x))$ $\displaystyle=$ $\displaystyle\mu f\circ\Psi(x)+\lambda g\circ\Psi(x)$ $\displaystyle=$ $\displaystyle\left(\mu\Psi^{\ast}(f)+\lambda\Psi^{\ast}(g)\right)(x),$

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 $y^{\prime}\in Y$, we have $y^{\prime}\notin\Psi(X)$. 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

 $f(y)=\begin{cases}v,&\mbox{if}\ y=y^{\prime},\\ 0,&\mbox{if}\ y\neq y^{\prime},y\in Y.\end{cases}$

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

Title $\Psi$ is surjective if and only if $\Psi^{\ast}$ is injective PsiIsSurjectiveIfAndOnlyIfPsiastIsInjective 2013-03-22 14:36:03 2013-03-22 14:36:03 matte (1858) matte (1858) 6 matte (1858) Theorem msc 03-00