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The following proposition is from Herrlich and Strecker's Category theory (2nd ed.):

29.1    PROPOSITION

Let $C \;\mathop{\rightrightarrows}\limits^f_g \;D$ be a pair of $\mathscr{A}$-morphisms [in some category $\mathscr{A}$]. Then the following are equivalent:

  1. $f = g$.
  2. For each $\mathscr{A}$-object $A$, $\hom(A, -)(f) = \hom(A, -)(g)$.

Proof: Clearly (1) implies (2). If (2) holds, then $ f = f\;{\scriptstyle\circ}\;1_C = \hom(C, f)(1_C) = \hom(C, g)(1_C) = g\;{\scriptstyle\circ}\;1_C = g.\;\;\;\;\square $

I think I understand both the theorem's statement and its proof, and yet the whole thing makes no sense to me. First, the $1 \Rightarrow 2$ implication is at once too banal to merit notice, and unnecessarily weak, since $f = g$ in fact implies $F(f) = F(g)$ for every functor $F$.

Second, the $2 \Rightarrow 1$ implication is more interesting, but I don't understand why its antecedent was made so strong.

Why isn't enough to require simply that $\hom(C, -)(f) = \hom(C, -)(g)$?

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    To give an example, one proves that in $\mathbf{Set}$, a function $f\colon X\to Y$ is one-to-one if and only if for all $Z$ and for all $g,h\colon Z\to X$, $fg=fh\Rightarrow g=h$, if and only if for all $g,h\colon X\to X$, $fg=fh\Rightarrow g=h$, if and only if for every one element set $W$, for all $g,h\colon W\to X$, $fg=fh\Rightarrow g=h$. That is, the condition is equivalent to something happening "universally", but in fact it's also equivalent to it happening for a specific "test object".2012-02-01

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You're right, it is sufficient to require just that $\hom(C, -)(f)$ and $\hom(C, -)(g)$ are the same function $\text{Hom}_{\cal A}(C,C)\to\text{Hom}_{\cal A}(C,D)$. As the cited proof says, $\hom(C, -)(f)$ sends $1_C$ to $f\circ 1_C=f$ and $\hom(C, -)(g)$ sends $1_C$ to $g\circ 1_C=g$, so if they are the same function, we must have $f=g$.

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    Well, regarding aside on Bernoulli, finally "the penny dropped"; now I see that the form "$p^x(1-p)^{1-x}$ for $x\in\{0, 1\}$" can be defended as one that shows it as the $n = 1$ case of the binomial distribution $B(n, p)$ (PMF = ${n\choose x}p^x(1-p)^{n-x}, x\in\{0,...,n\}$). Still, I found the $p^x(1-p)^{1-x}$ in a textbook introduction to the Bernoulli, without any mention, and in fact, well ahead of, the introduction to the binomial. At the very least, examples like these are evidence of obliviousness to how much confusion such *seemingly gratuitous* complexity can generate.2012-02-02