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$\DeclareMathOperator{\Frac}{Frac}$How does one prove the uniqueness of a homomorphism? My case is about the Universal Property of Fraction Fields. Let $R$ be an integral domain, $\iota: R\to \Frac(R)$ is the inclusion homomorphism taking $r$ to $r/1$. I have the homomorphism $\psi: \Frac(R)\to F$, where $\Frac(R)$ is an injective ring homomorphism defined as $\psi(a/b) = \phi(a)\phi^{-1}(b)$, where $\phi: R\to F$ is another injective (not explicitly defined) homomorphism. ($R$ is a ring). It is known that $\psi$ is injective and that it must satisfy $\psi\circ\iota=\phi$ (and it does). But how does one prove that $\psi$ is actually unique?

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    do you want this special case or a more general category theoretical one? I could do either if you want. – 2017-02-27
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    @ZelosMalum I don't yet know category theory, so a special case would do. – 2017-02-27

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Suppose $\chi\colon\operatorname{Frac}(R)\to F$ is another ring homomorphism such that $\chi\circ\iota=\phi$. Then for any $r\in R$, $$ \chi(r/1)=\chi\iota(r)=\phi(r)=\psi\iota(r)=\psi(r/1). $$

Also, for any $r\neq 0$ in $R$, $$ \chi(1/r)=\chi((r/1)^{-1})=\chi(r/1)^{-1}=\psi(r/1)^{-1}=\psi(1/r). $$

Then for $r\in R$, and $s\in R$ nonzero, $$ \chi(r/s)=\chi\left(\frac{r}{1}\cdot\frac{1}{s}\right)=\chi(r/1)\chi(1/s)=\psi(r/1)\psi(1/s)=\psi(r/s). $$

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    But how do we know that $\chi$ is not defined differently but just gives the same result? – 2017-02-27
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    @sequence Two functions $X\to Y$ are defined to be equal if they agree on every point of $X$. The above shows $\chi$ and $\psi$ agree on every element of $\operatorname{Frac}(R)$, hence $\chi=\psi$ be definition. – 2017-02-27
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    I thought the question of uniqueness assumes that if there are two functions $f$ and $g$ that output the same values, then they are different only if they are defined differently. But, after all, can there be a function defined differently which brings out the same values as some other function? So I just got confused by the formulation of the question. The question probably means that $\psi$ is unique only in the sense that $\psi\circ \iota = \phi$. However, again, how could it be not unique if $\psi\circ\iota = \phi$ if and only if, for all $r\in R$, $\psi\circ \iota(r) = \phi(r)$? @BenWest – 2017-02-27
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    @sequence The uniqueness statement means that if you fix $\iota$ and pick any $\phi\colon R\to F$, there is a unique ring map $\psi$ such that $\phi=\psi\circ\iota$. The explicit definition of $\psi$ shows one such map exists. The above answer shows any other ring homomorphism $\chi$ that satisfies the condition $\chi\circ\iota=\phi$ necessarily agrees with $\psi$ on every point of their common domain, hence **by definition**, they are equal functions. Even if you write down 2 functions by some rules which may look different at a first glance, as long as the values are the same on every point – 2017-02-27
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    the functions are equal. Technically a function $f\colon X\to Y$ is a special subset of $X\times Y$ given by ordered pairs $(x,y)$ such that for each $x\in X$, there is a unique pair of form $(x,y)\in X\times Y$, for some $y$. Two functions are equal if these sets are equal, which is what we are checking when we see if two functions agree at every point in their domain. – 2017-02-27
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    Can you please give an example of two functions that have the same sets $X\times Y$ but are defined differently? @BenWest – 2017-03-10
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    @sequence You could take something like $\frac{(x+3)(x-2)}{x-2}$ and $x+3$ as functions $(0,1)\to\mathbb{R}$, i.e, subsets of $(0,1)\times\mathbb{R}$. They're equal because they agree at every point in $(0,1)$. – 2017-03-10