Image of Composite Mapping

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Theorem

Let $f: S \to T$ and $g: R \to S$ be mappings.


Then:

$\operatorname{Im} \left({f \circ g}\right) = f \left[{\operatorname{Im} \left({g}\right)}\right]$

where:

$f \circ g$ is the composition of $g$ and $f$
$\operatorname{Im}$ denotes image
$f \left[{\cdot}\right]$ denotes taking image of a subset under $f$.


Corollary

Let $f: S \to T$ and $g: R \to S$ be mappings.


Then:

$\Img {f \circ g} \subseteq \Img f$


Proof

By definition of image, we have:

$\operatorname{Im} \left({f \circ g}\right) = \left\{{t \in T: \exists r \in R: f \circ g \left({r}\right) = t}\right\}$

and by definition of the image of a subset:

$f \left({\operatorname{Im} \left({g}\right)}\right) = \left\{{t \in T: \exists s \in \operatorname{Im} \left({g}\right): f \left({s}\right) = t}\right\}$

which, expanding what it means that $s \in \operatorname{Im} \left({g}\right)$, equals:

$f \left({\operatorname{Im} \left({g}\right)}\right) = \left\{{t \in T: \exists s \in S: \exists r \in R: g \left({r}\right) = s \land f \left({s}\right) = t}\right\}$

Now substituting $g \left({r}\right) = s$ in $f \left({s}\right) = t$, we obtain:

$f \left({\operatorname{Im} \left({g}\right)}\right) = \left\{{t \in T: \exists s \in S: \exists r \in R: f \left({g \left({r}\right)}\right) = t}\right\}$

which is seen to equal the expression for $\operatorname{Im} \left({f \circ g}\right)$ as soon as $S$ is non-empty.


The remaining case to be checked is thus if $S = \varnothing$.

From Null Relation is Mapping iff Domain is Empty Set, also $R = \varnothing$, so that $f$ and $g$ are empty mappings.

By Image of Empty Set is Empty Set, we conclude:

$\operatorname{Im} \left({f \circ g}\right) = \operatorname{Im} \left({g}\right) = \varnothing$

and also:

$f \left({\operatorname{Im} \left({g}\right)}\right) = \varnothing$

which together yield the desired equality.


Hence the result.

$\blacksquare$


Also see