# A new easy theorem in my draft

A new easy theorem in my draft.

Theorem $\mathrm{DOM} (g \circ f) \supseteq \mathrm{DOM} f$, $\mathrm{IM} (g \circ f) \supseteq \mathrm{IM} g$, $\mathrm{Dom} (g \circ f) \supseteq \mathrm{Dom} f$, $\mathrm{Im} (g \circ f) \supseteq \mathrm{Im} g$ for every composable morphisms $f$, $g$ of a category with restricted identities.

Proof $\mathcal{E}_{\mathcal{C}}^{Y, \mathrm{Dst} f} \circ \mathcal{E}_{\mathcal{C}}^{\mathrm{Dst} f, Y} \circ g \circ f = g \circ f \Leftarrow \mathcal{E}_{\mathcal{C}}^{Y, \mathrm{Dst} f} \circ \mathcal{E}_{\mathcal{C}}^{\mathrm{Dst} f, Y} \circ g = g$ and it implies $\mathrm{IM} (g \circ f) \supseteq \mathrm{IM} g$. The rest follows easily.

Corollary $\mathrm{dom} (g \circ f) \sqsubseteq \mathrm{dom} f$, $\mathrm{im} (g \circ f) \sqsubseteq \mathrm{im} g$ whenever $\mathrm{dom}$/$\mathrm{im}$ are defined.