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A Grothendieck topology by definition consists of sieves – what Johnstone calls a sifted coverage – whereas a Grothendieck pretopology in any non-trivial case will contain a non-sieve. (Recall that $\lbrace \textrm{id} : X \to X \rbrace$ is always a covering family for $X$, but it is a sieve if and only if there are no morphisms $Y \to X$ for any $Y \ne X$.) Thus, in any case of interest, no topology is a pretopology and no pretopology is a topology.

But siftedness is not the key difference between topologies and pretopologies. The key difference is saturation: as you are already aware, it is possible to add covering families to a pretopology without changing the category of sheaves. One can define the non-sifted analogue of a topology as a family of sinks satisfying the following axioms:

  • Any isomorphism constitutes a singleton covering family.

  • The composition of covering families is a covering family.

  • If a sink covering family factors through a covering familygiven sink, then it the sink is also a covering family.

One can show that every pretopology is contained in a unique such saturated coverage, and every saturated coverage contains a unique topology – just pick out the sieves!

Here's a reasonably "natural" example of a pair of pretopologies that generate the same topology. We consider the category $\textbf{Top}$ of topological spaces, or any full subcategory $\mathbf{T}$ thereof closed under pullbacks and open subsets.

  1. A sink $\lbrace f_i : U_i \to X \rbrace$ is covering if and only if each $f_i$ is open and a homeomorphism onto its image, and the union of the images is the whole of $X$.

  2. A sink $\lbrace f_i : Y_i \to X \rbrace$ is covering if and only if the induced map $f : \coprod_i Y_i \to X$ is a local homeomorphism.

Clearly, pretopology (1) is contained in pretopology (2). Conversely, given a covering family of type (2), we can obtain a covering family of type (1) by taking a suitable refinement. (For each point $y$ of $Y_i$, take an open neighbourhood $U_{i,j}$ that is mapped homeomorphically into $X$.) So the two pretopologies must generate the same topology.

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A Grothendieck topology by definition consists of sieves – what Johnstone calls a sifted coverage – whereas a Grothendieck pretopology in any non-trivial case will contain a non-sieve. (Recall that $\lbrace \textrm{id} : X \to X \rbrace$ is always a covering family for $X$, but it is a sieve if and only if there are no morphisms $Y \to X$ for any $Y \ne X$.) Thus, in any case of interest, no topology is a pretopology and no pretopology is a topology.

But siftedness is not the key difference between topologies and pretopologies. The key difference is saturation: as you are already aware, it is possible to add covering families to a pretopology without changing the category of sheaves. One can define the non-sifted analogue of a topology as a family of sinks satisfying the following axioms:

  • Any isomorphism constitutes a singleton covering family.

  • The composition of covering families is a covering family.

  • If a sink factors through a covering family, then it is also a covering family.

One can show that every pretopology is contained in a unique such saturated coverage, and every saturated coverage contains a unique topology – just pick out the sieves!

Here's a reasonably "natural" example of a pair of pretopologies that generate the same topology. We consider the category $\textbf{Top}$ of topological spaces, or any full subcategory $\mathbf{T}$ thereof closed under pullbacks and open subsets.

  1. A sink $\lbrace f_i : U_i \to X \rbrace$ is covering if and only if each $f_i$ is open and a homeomorphism onto its image, and the union of the images is the whole of $X$.

  2. A sink $\lbrace f_i : Y_i \to X \rbrace$ is covering if and only if the induced map $f : \coprod_i Y_i \to X$ is a local homeomorphism.

Clearly, pretopology (1) is contained in pretopology (2). Conversely, given a covering family of type (2), we can obtain a covering family of type (1) by taking a suitable refinement. (For each point $y$ of $Y_i$, take an open neighbourhood $U_{i,j}$ that is mapped homeomorphically into $X$.) So the two pretopologies must generate the same topology.