July 19 Information

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

Question:

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

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

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

Even if the domain of ${\displaystyle \phi }$ were required to be the entire codomain of ${\displaystyle f}$ 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 ${\displaystyle \phi _{1}:A\to C}$ and ${\displaystyle \phi _{2}:B\to D}$ such that ${\displaystyle g\circ \phi _{1}=\phi _{2}\circ f}$, 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) reply

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) reply