Exercise 28
If $X^{\circlearrowleft \alpha}$ is any object of $S^\circlearrowleft$ for which there exists an injective $S^\circlearrowleft$-map $f$ to some $Y^{\circlearrowleft \beta}$ where $\beta$ is in the subcategory of automorphisms, then $\alpha$ itself must be injective.
Notation and Definitions
- $S$ is the category of sets, with functions as morphisms ($S$-maps).
- $X^{\circlearrowleft \alpha}$ denotes a set $X$ equipped with an endomap $\alpha$.
- $S^\circlearrowleft$ is the category of endomaps of sets. An $S^\circlearrowleft$-map between objects $X^{\circlearrowleft \alpha}$ and $Y^{\circlearrowleft \beta}$ is an $S$-map $X \xrightarrow{f} Y$ which satisfies the equation $f \circ \alpha = \beta \circ f$.
- A map $X \xrightarrow{a} Y$ is injective iff for any maps $T \xrightarrow{x_1} X$ and $T \xrightarrow{x2} X$ (in the same category) if $a \circ x_1=a \circ x_2$ then $x_1=x_2$.
Partial Solution
If $T \xrightarrow{x_1} X$ and $T \xrightarrow{x2} X$ are $S$-maps such that $\alpha \circ x_1 = \alpha \circ x_2$, then $f \circ \alpha \circ x_1 = f \circ \alpha \circ x_2$. Since $f$ is an $S^\circlearrowleft$-map, $\beta \circ f \circ x_1 = \beta \circ f \circ x_2$. Composing boths sides with $\beta^{-1}$ ($\beta$ is an automorfism), we get $f \circ x_1 = f \circ x_2$.
If $x_1$ and $x_2$ were not only $S$-maps, but also $S^\circlearrowleft$-maps (for a given endomap $\tau$ on $T$), by the injectivity of $f$, we'd get $x_1=x_2$. However, this may not be the case. Therefore, my question is: how do I generalize this conclusion to all pairs of $S$-maps? I haven't found a way to do so, nor have I found any counterexamples. Should the statement of the problem be understood as making the stronger assumption that $f$ is injective as an $S$-map (in which case the previous proof suffices)?