Topology: A Categorical Approach, Exercise 0.3 d), e) 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 $[0, 1] \to (0, 1)$?
 A: 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.
A: It is indeed true that a monomorphism in $\mathbf{Top}$ is an injection, because this holds in $\mathbf{Set}$ and we can give any set the discrete in the domain or indiscrete topology in the comomain and make any function between sets a morphism in $\mathbf{Top}$ etc.
So in $\mathbf{Top}$ $f$ is a monomorphism iff $f$ is an injective continuous function, and $f$ is an epimorphism iff $f$ is a surjective continuous function.
A standard example of a continuous bijection $f:A \to B$ between spaces that is not a homeomorphism is to take any set $X$ of two or more points, where $A = (X, \mathscr{P}(X))$ (the discrete topology on $X$) and $B=(X,\{\emptyset,X\})$ (the indiscrete topology on $X$) and $f(x)=x$ for all $x \in X$.
Then this is not an isomorphism in $\mathbf{Top}$ (i.e. a homeomorphism) because $A$ is Hausdorff and $B$ is not or simply because $f$ is not an open map (a singleton in $A$ is open and its image in $B$ is not etc.).
In $\mathbf{Top}$ we can even have continuous bijections $f: X \to Y$ and $g: Y \to X$ and still $X$ and $Y$ being non-homeomorphic, see e.g. this old question on Mathoverflow.
Or more simply $X = [0,1]$ and $Y = (0,1)$ in the Euclidean topology both. One is compact, the other is not, so not homeomorphic (not isomorphic in $\mathbf{Top}$).
$X$ embeds into $Y$ via $f(x)=\frac{1}{2} + \frac{1}{3}x$ e.g. and $Y$ is even a subspace of $X$ so $i(x)=x$ will do.
