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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?

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Could you exhibit a homotopy of maps on non-Hausdorff spaces? –  user02138 Oct 6 '11 at 3:13
    
What's your proof? –  Joe Johnson 126 Oct 6 '11 at 4:30
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A real quick summary: "Lengthen" the intersections into line segments. On each sphere contract all the points where line segments intersect it into one point, so that if the sphere originally intersected $k$ other spheres, there will now be $k$ line segments emanating out of one point. We now essentially have a graph in which the vertices are $2$-spheres. Contract the line segments that make up the graph for as long as you can until there are no other contractible segments left. The result will be a wedge sum of $1$-spheres and $2$-spheres. –  Jonathan Gleason Oct 6 '11 at 22:22

1 Answer 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.

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So I think that at least part of the problem here is the ambiguity in how the question was phrased. I interpreted the question to speak of a space that is constructed by taking the disjoint union of finitely many $2$-sphere's, where a set is open iff it intersects each sphere in an open set, and then by identifying pairs of points on distinct spheres, equipping the resulting set with the quotient topology. Perhaps my interpretation was incorrect, but under that assumption, I do not believe this example is an example of such a space. –  Jonathan Gleason Oct 6 '11 at 22:17
    
In particular, from the way in which the question was worded, I did not think that, for example, equipping the disjoint union of two $2$-spheres with this ``exotic'' (i.e., not the topology that makes the disjoint union a coproduct) topology was allowed. –  Jonathan Gleason Oct 6 '11 at 22:19
    
But such a space is Hausdorff, so you've built in the assumption of Hausdorff-ness into your interpretation of the question. That is, the answer to your original question is "You've already used the assumption that the space is Hausdorff by assuming that it can be built in this way." –  Tom Church Oct 10 '11 at 7:57

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