# $\hom(C, -)(f) = \hom(C, -)(g)$ implies $f = g$

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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In fact, one could state this as an equivalence of 3 statements: the given statements 1 and 2, plus a third statement that only gives the equality you give. That would make it more in line with many such theorems: the functions are equal if and only if the two functions always behave the same, if and only if they behave the same with respect to a particular "test" object. –  Arturo Magidin Feb 1 '12 at 19:43
@ArturoMagidin: I'm afraid I would have found your proposal even more confusing than the original, for similar reasons. If, for example, a teacher filled the blackboard statements that are perfectly obvious to the students, such as "9 - 2 = 7 ... 3/2 = 0 + (-3)/(-2) ... 147 > -33 ...", the students would be as confused as ever, not because they disagree with the statements, but because the whole performance just makes no sense. Similarly, the statement "the functions are equal if and only if the two functions always behave the same" is, at this stage of the game, plain confusing. –  kjo Feb 1 '12 at 20:41
I'm trying to link this particular kind of phrasing to one that is often found in the literature in many contexts: one establishes that a certain property X holds if and only if Y holds "for all possible z", and that this holds if and only if Y holds "for a particular choice of z." My paraphrase was not meant to be formal or precise, merely indicative ("...if and only if for all A blah" and then showing this in turn is equvalent to "... if and only if for this specific choice of A blah"). –  Arturo Magidin Feb 1 '12 at 20:45
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". –  Arturo Magidin Feb 1 '12 at 20:48

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 $f=g$.
Thanks! I study math independently (I'm not a mathematician) and I keep running into this sort of unnecessary obfuscation all the time in mathematical writing (e.g. I just came across, in a statistics textbook, the mass function of the Bernoulli distribution described as "$f(x)=p^x(1−p)^{1−x}$ for $x \in \{ 0,1\}$"–who ever thinks about it this way???); but it takes time and mental effort to sort out the gratuitous obfuscation from the necessary complexity. It's all about "Gricean implicature": readers reasonably expect writers to be adults, and not engage in such adolescent shenanigans... –  kjo Feb 2 '12 at 0:55
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. –  kjo Feb 2 '12 at 21:46