Homotopy type of surface of revolution Let $X$ be a finite graph lying in a half-plane $P\subset\mathbb{R}^{3}$ and intersecting the edge of $P$ in a subset of the vertices of $X$. Describe the homotopy type of the surface of revolution obtained by rotating $X$ about the edge line of $P$.
Since distinct connected components of $X$ give rise to disjoint surfaces, I assumed $X$ is a connected graph. If $X$ has vertices on the edge of $P$, we can collapse edges of $X$ to those vertices. There are then two types of edges. The first type joins two distinct vertices. The second type is a loop. Rotating the first type gives a 2-sphere but what about the second type?
Am I right to say that the whole surface of revolution is homotopy equivalent to a wedge of $S^{1}$'s and $S^{2}$'s?
If $X$ has no vertices on the edge of $P$, we can collapse edges of $X$ to form a wedge of circles. What will be the homotopy type of the surface of revolution then?
 A: Any connected finite graph is made out of cycles and trees. All trees are contractible, and every chordless cycle is homotopy equivalent to $S^1$, all wedged at a point to which all of our trees contracted to. Concatenation of trees and chordless cycles makes up any finite connected graph, so every connected component $X_i$ of your graph $X$, being a connected finite graph, is just a wedge of $1 - \chi(X_1)$ circles (you can check that this is equivalent to the number of chordless cycles). So the full graph is homotopy equivalent to the following:
$$X \simeq \coprod^n_{i=1} X_i \simeq 
 \coprod^n_{i=1}\bigvee^{1-\chi({X_i})}_{j=0} S^1$$
(Where $\coprod$ denotes a disjoint union).
Since the surface of revolution is just equivalent to taking a cartesian product with $S^1$, we get that surface of revolution of $X$, (denote it $Y$) is equivalent to (supposing there are no disjoint vertices):
$$Y = S_1 \times X = S_1 \times \coprod^n_{i=1}\underbrace{S^1 \vee S^1 \vee \cdots \vee S^1}_{1-\chi({X_i})\text{ times}} = 
\coprod^n_{i=1}\underbrace{S^1 \times S^1 \vee S^1 \times S^1 \vee \cdots \vee S^1 \times S^1}_{1-\chi({X_i})\text{ times}}
$$
Since $S^1 \times S^1$ is a torus ($T^2$), so every connected component $X_i$ of $X$ corresponds to a wedge of $1-\chi({X_i})$ torii. As such, the surface of revolution of such graph, is just a disjoint union of different wedges of torii (and circles whenever there is a disjoint tree since $S^1 \times \{x\} = S^1$).
As such for a graph with $m$ disjoint trees and $n$ disjoint non-trees your space is a disjoint union of $m$ circles ($S^1$) and $n$ components which are wedges of $1-\chi(X_i)$ torii each (where $1-\chi(X_i)$ is a number of chordless cycles in a corresponding component).
