# domain of initial $f : X \rightarrow Y$ in Haus equipped with coarsest topology?

If $f:X\rightarrow Y$ is initial in category Top then it is easy to proof that

(!) the topology on $X$ is the set of preimages of open sets in $Y$.

Just construct topology $Z$ having the same underlying subset as $X$ and let the set of these preimages serve as topology on it. Then from $g:Z\rightarrow X$ with $x\mapsto x$ it is clear that $fg$ is continuous so the conclusion that $g$ is continuous can be made. Then we are ready. But now my question:

what if we do not work in $\textbf{Top}$ but in category $\textbf{Haus}$?

The constructed topology $Z$ does not have to be a Hausdorff space (or am I overlooking something here?) and if the fact that $f$ is initial in $\textbf{Haus}$ would work then it would justify the conclusion that $g$ can be recognized as an arrow in $\textbf{Haus}$.

Is there a way out? Or - even stronger - is statement (!) not true in $\textbf{Haus}$?

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A continuous $f: X \to Y$ is by definition initial iff [$(∀Z\in\mathbf{Top} ) (∀g: Z \to X)$ $g$ is continuous iff $f g$ is continuous]. In $\mathbf{Top}$ that's equivalent to $X$ having the initial topology induced by $f$. And it is also equivalent in $\mathbf{Haus}$ since (as you noted) initial morphisms in $\mathbf{Haus}$ are precisely the embeddings.

Let $f: X \to Y$, $Y$ Hausdorff. Then the initial topology on $X$ is Hausdorff if and only if $f$ is injective. If $f$ is not injective then initial topology on $X$ isn't even $T_0$ since points with the same image are not topologically distinguishable. On the other hand if $f$ is injective, then initial topology on $X$ is subspace topology and hence Hausdorff.

Let $f: X \to Y$ be initial in $\mathbf{Haus}$ (so $X$, $Y$ are Hausdorff). Then $f$ is injective.
Proof: Let $Z$ be some Hausdorff space, $∅ ≠ A ⊊ Z$ non-open. Let $f(x) = f(x')$, then let $g: Z \to X$ map $A$ to $x$ and $Z \setminus A$ to $x'$. Then $g$ is not continuous since $A$ is non-open and $X$ is Hausdorff. But $fg$ is constant so $f$ cannot be initial.

So if $f: X \to Y$ is initial in $\mathbf{Haus}$ then the initial topology on $X$ is Hausdorff by the proposition above and so $f$ is an embedding.

Note that the same proofs work for $T_0$, $T_1$, $T_3$, $T_{3\frac{1}{2}}$ and any full subcategory of $T_0$ spaces closed under subspaces and containing some non-discrete space.

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Initial topologies need not be Hausdorff. For example, for any set $X$, the initial topology induced by the unique map $X \to 1$ is Hausdorff if and only if $X$ has cardinality $\le 1$.

Worse, there need not be an initial topology among the Hausdorff ones. For example, suppose $X$ is an infinite set with the same cardinality as $\mathbb{R}$. Then, for every subset $S \subseteq X$, if $S \ne X$ and $S \ne \emptyset$, then there exists a Hausdorff topology on $X$ such that $S$ is not open. (If $S$ has cardinality strictly less than $X$, then topologise $X$ so that $X \cong \mathbb{R}$, and observe that every non-empty open subset of $\mathbb{R}$ has cardinality equal to $\mathbb{R}$; otherwise, choose a suitable bijection $X \cong \mathbb{R}$ so that $S$ is identified with $[0, 1] \subseteq \mathbb{R}$, which is not open.) Hence, the only topology on $X$ that is coarser than all the Hausdorff topologies on $X$ is the indiscrete topology on $X$, which is not Hausdorff.

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In Joy of Cats it is stated that in Haus initial arrows coincide with topological embeddings (pg 135). This can be proved by showing that initial arrows in Haus are injective. That is what I did, but this under the assumption that statement (!) is true if f:X->Y is initial in Haus. So my 'proof' was wrong. Can someone give me a real proof of the statement? –  drhab Jul 26 '13 at 11:56
I received answer on the question mentioned in my comment above. This after changing the comment into a real question posed on MathStack. –  drhab Jul 27 '13 at 8:14