A common mistake people make is to think that a continuous bijection is a homeomorphism. This is a reasonable mistake. A bijection is an isomorphism of sets. A bijective homomorphism of groups is an isomorphism of groups. In most algebraic settings I can think of this pattern holds. But it is not true of topological spaces.
Here are the examples I first came up with when I first read Munkres. Intuitively, tearing is not continuous, and the inverse of tearing is gluing back together. I first thought of doing this with a disk, but the idea is the same if you do it with an interval. To be explicit, define ![f \colon [0,1) \cup [2,3] \to [0,2]](https://s0.wp.com/latex.php?latex=f+%5Ccolon+%5B0%2C1%29+%5Ccup+%5B2%2C3%5D+%5Cto+%5B0%2C2%5D&bg=060f29&fg=f2f2f2&s=0&c=20201002)
by
.
This is clearly a continuous bijection, but a disconnected space is not homeomorphic to a connected space. An inverse function of 
exists, but it is not continuous.
A minimal version of this counter example can be found by considering two point spaces. Consider the set 
, and let 
and 
denote the discrete and indiscrete topologies on 
respectively. Then the identity function 
is continuous when you give topologies like this: 
. The inverse of this function is itself, but when you switch the topologies, this is not continuous.
This makes me want to think about adjunctions. There is a forgetful functor 
which takes a topological space and just produces its underlying set. This functor has both a left and right adjoint. One takes a set and produces the discrete space with that set, and the other produces the indiscrete space with that set, lets call them 
and 
respectively. Both functors do the same thing on morphisms, nothing. You can figure out which one is left and which is right pretty quick with some basic point-set topology knowledge. Every function into an indiscrete space, and every function out of a discrete space, is automatically continuous. So we get the natural isomorphisms
.
Notice that 
and 
for any set 
. Then we get
.
This is funny to me because this isomorphism relates the identity function on to the counterexample we saw earlier, which is not an isomorphism. Without time to think about it, if you told me you had a functor with a left and right adjoint, and 
, I would have guessed that the natural isomorphisms would preserve isomorphisms, or at least that it would send the identity map to an isomorphism, but no!
Joe Moeller 
> In most algebraic settings I can think of this pattern holds.
For algebras of any Lawvere theory – algebras in the category Set, that is – a morphism that’s a bijection on the underlying sets is an isomorphism. Show this!
I recently made a mistake in claiming that any order-preserving bijection between posets had an order-preserving inverse. Dumb! Find a counterexample. This can be seen as a proof that posets are not the algebras of any Lawvere theory. But the moral is: _relations don’t work like operations!_
An algebra of a Lawvere theory is a product preserving functor with domain the theory. A morphism between algebras is a natural transformation. If the first component of such a morphism is a bijection, then so is every other component. Every component being invertible is the same as the whole thing being invertible. So a morphism which is a bijection is an isomorphism.
For the counterexample to the poset case, I think the same idea of the counterexample I gave for topological spaces should work. The identity-on-elements map
, where the first is the discrete (is that what its called?) poset, and the second is a linear order, is order preserving, but the inverse function is not since it breaks the one nontrivial relation.
Hmm, maybe you can turn on the markup language on your blog so things like _this_ will show up in italics, etc. It was pretty easy to do in my own (free) WordPress blog by going to a page for adjusting settings.