Suppose $(E,m)$ and $(E'm)$ are both equalizers of $f$ and $g$ show that there is an isomorphism $k\colon E\to E'$

Question

Suppose $f,g\colon X\to Y$ are functions. Then there is a set $E$ and a function $m\colon E\to X$ with the following property: $f\circ m=g\circ m$, and for any other function $h \colon F \to X$ with $f \circ h = g \circ h$ there is a unique function $k\colon F\to E$ such that $m\circ k=h$. We call $(E,m)$ an equalizer of $f$ and $g$.

What I have so far,

Suppose there is a function $i\colon E\to E'$

I am trying to see what composition of function, from those given in the definition, will be such that

$$i \circ \text{that function}=I_E\quad\text{and}\quad\text{that function} \circ i=I_{E'}$$

The problem is, there is no function(s) or composition thereof that arrows to $E$ and its counterpart to $E'$.

Any ideas?

Answer 1

As mentioned in the comments by azif00 it is not exactly clear if you want to show that $(E,m)$ exists or that it is unique up to (unique) isomorphism. As the existence is only a (simple) set-theoretic matter I will talk about uniqueness. What follows is the usual argument for showing that universal objects (i.e. objects defined by universal properties) are unique up to unique isomorphism.

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Let $f,g\colon X\to Y$ be two set-functions between set $X,Y$ (so we are working in $\sf Set$). The equalizer $\operatorname{eq}(f,g)$ of $f$ and $g$ is characterized as a pair $(\operatorname{eq}(f,g),m)=:(E,m)$ such that $f\circ m=g\circ m$ (I fixed a small typo there in your questions) and such that for every other pair $(E',m')$ with $f\circ m'=g\circ m'$ there is a unique set-function $k\colon E'\to E$ with $m\circ k=m'$. I decided to state the universal property again as your (original) version is slightly confusing and hides the crucial fact.

So, let us take two pairs $(E,m),(E',m')$ which both satisfy the aforegoing universal property. Using this property both ways around (so one time with $(E,m)$ as the set-function-inducing-pair and one time with $(E',m')$, but both times with the other as second object) we obtain unique set-functions:

$$k\colon E'\to E,\quad k'\colon E\to E'$$

We note, however, that we can use the universal property on the pair $(E,m)$ in both places! This gives us a unique set-function $i\colon E\to E$ such that $m\circ i=m$. But there already is such a set-function, namely the identity $\operatorname{id}_E\colon E\to E$, and by uniqueness $i=\operatorname{id}_E$. A similar argument may be applied to the pair $(E',m')$.

Now note that the compositions $k\circ k'\colon E\to E$ and $k'\circ k\colon E'\to E'$ both satisfy the very same properties as the corresponding identites on $E$ and $E'$, respectively, do. Hence, by uniqueness again, we have $k\circ k'=\operatorname{id}_E$ and $k'\circ k=\operatorname{id}_{E'}$. So, $k\colon E'\to E$, in particular, is a uniquely defined bijective set-function; that is an isomorphism in $\sf Set$.

Answer 2

Consider the category $\mathscr C$ that has

• as objects, pairs $(F,h)$ where $F$ is a set and $h$ is a function from $F$ to $X$ such that $f \circ h = g \circ h$; and
• a morphism $k$ from $(F_1,h_1)$ to $(F_2,h_2)$ is a function $k \colon F_1 \to F_2$ such that $h_1 = h_2 \circ k$.

Then note that an equalizer of $f$ and $g$ is an object $(E,m)$ in $\mathscr C$ such that for any other object $(F,h)$ in $\mathscr C$ there is a unique morphism $k$ from $(F,h)$ to $(E,m)$. In other words, $(E,m)$ is a terminal object in $\mathscr C$. Can you conclude from here?