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July 19

Embedding, monomorphism

By an "embedding" of a first given function $f$ in a second given function $g\text{ (}$or a "monomorphism" from a first given function $f$ to a second given function $g),$ I mean, as expected, a one-to-one mapping $\varphi$ from the image of $f$ to the domain of $g,$ satisfying$:g\circ \varphi =\varphi \circ f.$

Question:

Given a first function $f$ "embedded" in a second given function $g,$ and given a third function $h$ satisfying that the composition $h\circ f$ has a one-to-one mapping from the image of this composition to the domain of the second composition $h\circ g,$ can the first composition $h\circ f$ always be embedded in the second composition $h\circ g?$

2A06:C701:7471:3000:39AA:1A85:25C2:975B (talk) 19:59, 19 July 2023 (UTC)

Your definition of "embedding" doesn't really make sense for functions with different domains and codomains. If $f:A\to B$ and $g:C\to D$, then with your definition, an embedding would be a one-to-one map $\varphi :f(A)\to C$ such that $g\circ \varphi =\varphi \circ \overline{f}$ (where $\overline{f}$ is just $f$ but with a codomain of $f(A)$), and by comparing domains and codomains, that would imply that $f(A)=A$ and $C=D$.

Even if the domain of $\varphi$ were required to be the entire codomain of $f$ instead of just the image, the definition would still be too restrictive (though it would work if one were dealing only with endofunctions).

Instead, an embedding should be a pair of one-to-one maps ${\varphi }_1:A\to C$ and ${\varphi }_2:B\to D$ such that $g\circ {\varphi }_1={\varphi }_2\circ f$, i.e., a morphism in the arrow category of the category of sets (which is a special case of a comma category).

GeoffreyT2000 (talk) 21:01, 19 July 2023 (UTC)

Correct. I've just corrected my question in the following thread. See below. 2A06:C701:7471:3000:39AA:1A85:25C2:975B (talk) 17:03, 20 July 2023 (UTC)