Image of a composition

(1) $\operatorname{Im}(g\circ f)=g(\operatorname{Im} f)\subseteq g(F)=\operatorname{Im} g$.

(2) Equality $\Leftarrow$. Assume $\operatorname{Im} f+\ker g=F$ and let $w\in\operatorname{Im} g$, say $w=g(y)$. Write $y=y_f+k$ with $y_f\in\operatorname{Im} f$ and $k\in\ker g$. Then $w=g(y_f)+g(k)=g(y_f)\in g(\operatorname{Im} f)$.

Equality $\Rightarrow$. Assume $g(\operatorname{Im} f)=g(F)$ and let $y\in F$. Then $g(y)=g(y_f)$ for some $y_f\in\operatorname{Im} f$, so $k:=y-y_f\in\ker g$ and $y=y_f+k\in\operatorname{Im} f+\ker g$.

Surjective $f$. If $\operatorname{Im} f=F$, then $\operatorname{Im} f+\ker g=F$, so $\operatorname{Im}(g\circ f)=\operatorname{Im} g$.

(3) Counterexample. Take $f:\mathbb{R}\to\mathbb{R}$, $f=0$, and $g=\mathrm{id}_{\mathbb{R}}$ (which is both injective and surjective). Then $\operatorname{Im}(g\circ f)=\{0\}\neq\mathbb{R}=\operatorname{Im} g$. Here $\operatorname{Im} f+\ker g=\{0\}\neq\mathbb{R}$, as predicted by (2).

Conclusion. In general $\operatorname{Im}(g\circ f)\subseteq\operatorname{Im} g$, with equality exactly when $\operatorname{Im} f+\ker g=F$ (in particular when $f$ is surjective), and not merely when $g$ is surjective or injective.