Exercise:
My attempt:
I am not able to find examples for either d) or e), but I have some theories about what I might be looking for.
Regarding d), in $\mathsf{Top}$ the morphisms are continuous functions between topological spaces. I need a continuous function $f \colon X \to Y$ that is left- and right-cancellative yet not a homeomorphism. Since left- and right-cancellative means injective and surjective (for functions, at least), such a function $f$ will necessarily be bijective. Therefore I am looking for a continuous bijection $f \colon X \to Y$ such that $f^{-1}$ is not continuous.
Regarding e), a comment on MathOverflow said I could consider $(0, 1)$ and $[0, 1]$ from $\mathsf{Top}$, but the level of discussion there is a bit above my pay grade. I suppose the morphism between the two was considered obvious, but I don't know what it would be. I figure that I need an injective, continuous function each way, such that there cannot be a continuous bijection each way.
I appreciate any help.
Edit:
Now I am confused regarding e). Since continuity preserves compactness, doesn't that mean there is no morphism of any kind $[0, 1] \to (0, 1)$?
Best Answer
For d) think of any topological space $X$ with a topology $\tau$ that is not indiscrete then the identity function $id:(X,\tau)\to (X,I(X))$ satisfies the conditions.
For e) notice that any embedding $[0,1]\to (0,1)$ (for example the lineal bijection between $[0,1]$ and $[\frac{1}{4},\frac{1}{2}]$) and the inclusion $(0,1)\hookrightarrow[0,1]$ are two monomorphisms but the sets are clearly not homeomorphic.