Image of Composite Mapping

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, and the $f$ signifies taking image of a subset under $f$.

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 Mapping to 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.

Also see

 * Image and Preimage of Composite Relation