Proposition 1   The equalizer of a pair of parallel morphisms $f,g:A\to B$ in a category $\mathcal{C}$ is an inverse limit.

Proof. We need to find a category $\mathcal{I}$ and a functor $F:\mathcal{I}\to \mathcal{C}$ such that the equalizer of $f,g$ is the limit of functor $F$ The idea behind finding $\mathcal{I}$ is to look at the diagram of a pair of parallel morphisms, $$\xymatrix@+=3pc{A \ar@<0.5ex>[r]^f \ar@<-0.5ex>[r]_g & B} $$ and construct $\mathcal{I}$ based on the diagram. Thus, let $\mathcal{I}$ be the category consisting of two objects $a,b$ and four morphisms $1_a,1_b, r, s$ such that $r,s\in \hom(a,b)$ Define $F$ to be the functor such that $F(a)=A, F(b)=B, F(r)=f$ and $F(s)=g$

Suppose $(L,\tau)$ is the limit of $F$ Identify the constant functor $L$ with its value the object $X$ in $\mathcal{C}$ and the natural transformation $\tau$ a pair of morphisms $i:X\to F(a)$ and $j:X\to F(b)$ in $\mathcal{C}$ such that $$\xymatrix@+=3pc{X \ar[r]^-i & F(a) \ar[r]^{F(r)} & F(b)} = \xymatrix@+=3pc{X \ar[r]^-j & F(b)} = \xymatrix@+=3pc{X \ar[r]^-i & F(a) \ar[r]^{F(s)} & F(b)}.$$ This is the same as $$\xymatrix@+=3pc{X \ar[r]^-i & A \ar[r]^{f} & B} = \xymatrix@+=3pc{X \ar[r]^-i & A \ar[r]^{g} & B}.$$ So $i:X\to A$ equalizes $f$ and $g$

Suppose $(Y,\lbrace m:Y\to F(a), n:Y\to F(b)\rbrace)$ is another pair such that $$\xymatrix@+=3pc{Y \ar[r]^-m & F(a) \ar[r]^{F(r)} & F(b)} = \xymatrix@+=3pc{Y \ar[r]^-n & F(b)} = \xymatrix@+=3pc{Y \ar[r]^-m & F(a) \ar[r]^{F(s)} & F(b)}.$$ Then there is a unique morphism $k:Y\to X$ such that $$\xymatrix@+=3pc{Y \ar[r]^-k & X \ar[r]^i & F(a)} = \xymatrix@+=3pc{Y \ar[r]^-m & F(a)}.$$ This is the same as saying that whenever $m:Y\to A$ equalizes $f$ and $g$ there is a unique morphism $k:Y\to X$ such that $$\xymatrix@+=3pc{Y \ar[r]^-k & X \ar[r]^i & A} = \xymatrix@+=3pc{Y \ar[r]^-m & A},$$ which is exactly the statement that $i:X\to A$ is the equalizer of $f$ and $g$