**Question**: Consider the proper class of all $n$-dimensional smooth manifolds. If we take the equivalence classes where two manifolds are identified if there exists a diffeomorphism between them, is this collection of equivalence classes **a set**?

*Remark*: I do not assume my manifolds to be Hausdorff nor second countable. If the answer depends on those assertions (*Edit*: it most definitely does) I would like to hear about the difference.

*Remark 2*: As Omar pointed out, there may be a problem with the various long lines. To make the question slightly more tractable, I would be (mostly) satisfied if there is a statement even ignoring the smooth structure and consider the case of topological manifolds and homeomorphisms.

**Motivation**: for something that I am working on I need to consider the collection of all $n$-dimensional smooth manifolds satisfying "property $X$". Unfortunately property $X$ is diffeomorphism invariant, so most definitely this collection is not a set, which invalidates many constructions (I want to build a manifold out of this collection; if the collection is a proper class then even on the set level the thing that I constructed will be a proper class, instead of a set) or at least forces me to rethink how this constructions ought to go. Fortunately for my argument it suffices that I have one object in each diffeomorphism class in my collection.

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