$\mathrm{\Psi }$ 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 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 $f\ne 0$. ∎

Title	$\mathrm{\Psi }$ is surjective if and only if ${\mathrm{\Psi }}^{\ast }$ is injective
Canonical name	PsiIsSurjectiveIfAndOnlyIfPsiastIsInjective
Date of creation	2013-03-22 14:36:03
Last modified on	2013-03-22 14:36:03
Owner	matte (1858)
Last modified by	matte (1858)
Numerical id	6
Author	matte (1858)
Entry type	Theorem
Classification	msc 03-00