Prove that a topological space equipped with a delta-complex structure is Hausdorff Suppose that $X$ is a topological space equipped with a delta-complex structure. Prove that $X$ is Hausdorff.
I can actually "see" why it should be Hausdorff after the hint from Hatcher asks to "think of a delta-complex X as a quotient space of a collection of disjoint simplices".
I would appreciate if somebody can write out a full formal argument showing this proof. I'm trying to start from the definition of Hausdorff but I don't know how to carry on.
 A: $\newcommand{\ra} [1] {\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex}
 \newcommand{\da} [1] {\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} $As you are using Hatcher's book, you will find in the appendix a section about CW-complexes, together with a proof that they are $T_4$. If you compare the inductive construction of a CW-complex with Hatcher's definition of a $\Delta$-complex, you will see they are pretty similar.
There is a major difference, though. The inductive process of building a CW-complex yields a quotient of a disjoint union of balls, whereas we would like to have a cell-complex structure on a given set $Y$. So we should have characteristic maps $\Psi_\alpha:D_\alpha^n\to Y$ which are injective on the interior, such that each point of $Y$ is in the image of exactly on such interior. We can then define the topology inductively on $Y$. Define $Y^n$ as the union of the images of $\Psi_\alpha$ up to dimension $n$. We assume that the restriction of each $\Psi_\alpha$ on the boundary of $D_\alpha^n$ is a continuous map to $Y^{n-1}$. We will show by induction over $n$ that $Y$ is homeomorphic to the quotient space of balls induced by these maps. It starts with $X^0$ which is the disjoint union of points $D^0$ and is homeomorphic to the discrete space $Y^0$. Assume now that $\Psi^n$ is a homeomorphism from $X^n$ to $Y^n$. Then the first arrow in the diagram is a homeomorphism.
$\large
\begin{array}{c}
X^n\sqcup\bigsqcup_\alpha D_\alpha^{n+1} & \ra{\Psi^n\sqcup\text{Id}} & Y^n\sqcup\bigsqcup_\alpha D_\alpha^{n+1}\\
\da{q} & & \da{\text{Id}\sqcup\bigsqcup\Psi_\alpha}\\
X^{n+1} & \ra{\Psi^{n+1}} & Y^{n+1}
\end{array}$
We will define $X^{n+1}$ as the adjunction space obtained by gluing the $D_\alpha^{n+1}$ along their boundaries to $X^n$ via $\phi(x)=(\Psi^n)^{-1}(\Psi_\alpha(x))$. So an identification takes place iff the images along the upper right path are the same. Since we topologize $Y^{n+1}$ as a quotient space, we get a homeomorphism on the bottom arrow.
This shows that $Y$ can also be considered a CW-complex although, as a set, it is not a disjoint union of balls.
So how can this be applied to a $\Delta$-complex? Well, take as $\Psi_\alpha$ the maps $\sigma_\alpha$. Property (ii) says that the restrictions are continous maps into the skeleton which exists by then, and property (iii) assures that the topology is the right one.
But a direct proof isn't difficult and would follow the idea of Proposition A.3 in the appendix. Assume by in induction that disjoint open neighborhoods $N^n(a)$ and $N^n(b)$ in the $n$-skeleton have been constructed. Then for a simplex $\Delta^{n+1}_\alpha$ there are four cases:


*

*$a$ and $b$ are in the interior of $\Delta^{n+1}_\alpha$: By Hausdorff there are disjoint open neighborhoods.

*$\sigma_\alpha^{-1}(a)$ is in the boundary and $b$ in the interior: By normality there are disjoint open $U,V$ with $\overline{\sigma_\alpha^{-1}(N^n(a))}\subset U$ and $b\in V$.

*$\sigma_\alpha^{-1}(a)$ and $\sigma_\alpha^{-1}(b)$ are both in the boundary: Then $\sigma_\alpha^{-1}(N^n(a))$ and $\sigma_\alpha^{-1}(N^n(b))$ can be "thickened" a bit towards the barycenter and remain disjoint.

*If one of $\sigma_\alpha^{-1}(a)$, $\sigma_\alpha^{-1}(b)$ is empty (Note that in this case also the corresponding $\sigma_\alpha^{-1}(N^n(-))$ is empty due to the way we construct the neighborhoods), then we can take the union with int($\Delta^{n+1}_\alpha$). If both are empty, there is nothing to do.


Doing this for every $n+1$-simplex yields the $N^{n+1}$'s, disjoint neighborhoods in the $n+1$-skeleton.
