Consider the following recursive definition of the factorial function. \[ FAC = \lambda n.\;IF \left(=\;n\;0\right)\;1\;\left(*\;n\;\left(FAC\;\left(-\;n\;1\right)\right)\right) \nonumber \] The definition relies on the ability to name a $\lambda$-abstraction and then to refer to this name inside the $\lambda$-abstraction itself. No such facility is provided by the $\lambda$-calculus. $\beta$-abstraction is applying $\beta$-reduction backwards to introduce new $\lambda$-abstractions, thus $+\;4\;1\leftarrow \left(\lambda x.\;+\;x\;1\right)\; 4$. By $\beta$-abstraction on $FAC$, its definition can be written \[ FAC = \left(\lambda fac.\;\left(\lambda n.\;IF\left(=\;n\;0\right)\;1\;\left(*\;n\;\left(fac\;\left(-\;n\;1\right)\right)\right)\right)\right) FAC \nonumber \] This definition has taken the form $FAC = g\;FAC$ where $g = \left(\lambda fac.\;\left(\lambda n.\;IF\left(=\;n\;0\right)\;1\;\left(*\;n\;\left(fac\;\left(-\;n\;1\right)\right)\right)\right)\right)$ is without recursion. We see also that $FAC$ is a fixed point ("fixpoint") of $g$. It is clear this fixed point can only depend on $g$ so supposing there were a function $Y$ which takes a function and delivers a fixpoint of the function as the result, we'd have $FAC = Y\;g = g\;(Y\;g)$. Under the assumption such a function exists, in order to build confidence this definition of $FAC$ works, we will try to compute $FAC\;1$. Recall \[ \begin{eqnarray} &FAC& = Y\;g \nonumber \\ &g& = \lambda fac.\;\left(\lambda n.\;IF\left(=\;n\;0\right)\;1\;\left(*\;n\;\left(fac\;\left(-\;n\;1\right)\right)\right)\right) \nonumber \end{eqnarray} \] So, \[ \begin{eqnarray} FAC\;1 &\rightarrow& (Y\;g)\; 1 \nonumber \\ &\rightarrow& (g\;(Y\;g))\;1 \nonumber \\ &\rightarrow& (\left(\lambda fac.\;\left(\lambda n.\;IF\left(=\;n\;0\right)\;1\;\left(*\;n\;\left(fac\;\left(-\;n\;1\right)\right)\right)\right)\right) (Y\;g))\; 1 \nonumber \\ &\rightarrow& \left(\lambda n.\;IF\left(=\;n\;0\right)\;1\;\left(*\;n\;\left(\left(Y\;g\right)\;\left(-\;n\;1\right)\right)\right)\right)\; 1 \nonumber \\ &\rightarrow& *\;1\;\left(\left(Y\;g\right)\;0\right) \nonumber \\ &\rightarrow& *\;1\;\left(\left(g\;\left(Y\;g\right)\right)\;0\right) \nonumber \\ &\rightarrow& *\;1\;\left(\left(\left(\lambda fac.\;\left(\lambda n.\;IF\left(=\;n\;0\right)\;1\;\left(*\;n\;\left(fac\;\left(-\;n\;1\right)\right)\right)\right)\right)\;\left(Y\;g\right)\right)\;0\right) \nonumber \\ &\rightarrow& *\;1\;\left(\left(\lambda n.\;IF\left(=\;n\;0\right)\;1\;\left(*\;n\;\left(\left(Y\;g\right)\;\left(-\;n\;1\right)\right)\right)\right)\;0\right) \nonumber \\ &\rightarrow& *\;1\;1 \nonumber \\ &=& 1 \nonumber \end{eqnarray} \]

The $Y$ combinator of the $\lambda$-calculus is defined as the $\lambda$-term $Y = \lambda f.\;\left(\lambda x.\;f\;\left(x\;x\right)\right)\left(\lambda x.\;f\;\left(x\;x\right)\right)$. $\beta$ reduction of this term applied to an arbitrary function $g$ proceeds like this: \[ \begin{eqnarray} Y\;g &\rightarrow& \left(\lambda f.\;\left(\lambda x.\;f\;\left(x\;x\right)\right) \left(\lambda x.\;f\;\left(x\;x\right)\right)\right)\;g \nonumber \\ &\rightarrow& \left(\lambda x.\;g\;\left(x\;x\right)\right) \left(\lambda x.\;g\;\left(x\;x\right)\right) \nonumber \\ &\rightarrow& g\;\left(\left(\lambda x.\;g\;\left(x\;x\right)\right)\;\left(\lambda x.\;g\;\left(x\;x\right)\right)\right) \nonumber \\ &=& g\;\left(Y\;g\right) \end{eqnarray} \] The application of this term has produced a fixpoint of $g$. That is, we are satisfied that this term will serve as a definition for $Y$ having the property we need and call it the "fixpoint combinator".

In the untyped $\lambda$-calculus, $Y$ can be defined and that is sufficient for expressing all the functions that can be computed without having to add a special construction to get recursive functions. In typed $\lambda$-calculus, $Y$ cannot be defined as the term $\lambda x.\;f\;(x\;x)$ does not have a finite type. Thus, when implementing recursion in a functional programming language it is usual to implement $Y$ as a built-in function with the reduction rule $Y\;g \rightarrow g\;(Y\;g)$ or, in a strict language, $(Y\; g)\;x \rightarrow (g\;(Y\;g))\;x$ to avoid infinite recursion.

For an OCaml like language, the idea then is to introduce a built-in constant $\mathbf{Y}$ and to denote the function defined by $\mathbf{let\;rec}\;f\;x = e$ as $\mathbf{Y}(\mathbf{fun}\;f\;x \rightarrow e)$. Intuitivly, $\mathbf{Y}$ is a fixpoint operator that associates a functional $F$ of type $\alpha \rightarrow \beta$ with a fixpoint $Y(F)$ of type $\alpha \rightarrow \beta \rightarrow \alpha$, that is, a value having the property $\mathbf{Y}\;F = F\;\left(\mathbf{Y}\;F\right)$. The relevant deduction rules involving this constant are: \[ \begin{equation} \frac{\vdash f\;(Y\;f)\;x \Rightarrow v} {\vdash (Y\;f)\;x \Rightarrow v} \tag{App-rec} \end{equation} \] \[ \begin{equation} \frac{\vdash e_{2}\left[Y(\mathbf{fun}\;f\;x \rightarrow e_{1})/f\right] \Rightarrow v} {\vdash \mathbf{let\;rec}\;f\;x=e_{1}\;\mathbf{in}\;e_{2} \Rightarrow v} \nonumber \tag {Let-rec} \end{equation} \]

References:

[1] The Implementation of Functional Programming Languages,Simon Peyton Jones, 1987.

[2] The Functional Approach to Programming, Guy Cousineau, Michel Mauny, 1998.