Question in Hatcher

Exercise 0.21 of Hatcher's Algebraic Topology reads:

If $X$ is a connected Hausdorff space that is the union of a finite number of $2$-spheres, any two of which intersect in at most one point, show that $X$ is homotopy equivalent to a wedge sum of $S^1$'s and $S^2$'s.

I believe I came up with a solution to this, but nowhere did I use the assumption of "Hausdorff". Is this really a necessary assumption? Where would you use the $T_2$ condition in a proof like this? It seems that a union of $2$-spheres would have to be Hausdorff . . .

What am I missing?

Solution 1:

Consider the disjoint union $X = S^2 \coprod S^2$ of two 2-spheres, topologized so that every neighborhood of $p$ on the first sphere contains $q$ on the second sphere and vice versa. (We have "glued $p$ to $q$", but by putting them infinitesimally close together, rather than identifying them as we would have to if we wanted to stay Hausdorff.) This space is connected and it is the union of two 2-spheres which meet in at most one point (since they meet in zero points).

In this case I would guess that the map $X$ to $S^2 \vee S^2$ actually is a homotopy equivalence — you can define a candidate for the inverse by taking the first sphere to $S^2$ (sending the basepoint to $p$) and taking the second sphere minus the basepoint to $S^2 - q$ (note that this really is continuous). However it does provide an example of how our intuition can go wrong if we allow non-Hausdorff spaces.