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.

Leave a Reply