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Suppose we have continuous function $f : X \to X$ that sends the basepoint of $X$ to itself, viz. $f(x_0) = x_0$ where $x_0$ is the basepoint of $X$. Recall the definition of the mapping torus $T_f$ of a space $X$. It is defined to be $X \times I$ mod by the equivalence where we identify $(x,0) \cong (f(x),1)$.

Now I have problem 1.2.11 of Hatcher as my assignment problem where I am suppose to compute the fundamental groups of two spaces using the induced map

$$ f_\ast : \pi_1(X) \longrightarrow \pi_1(X).$$

The first of these, when $X=S^1 \vee S^1$ I have computed the mapping torus $T_f$. I did it by first observing that because $f$ is basepoint preserving, we have that $(x_0,0) \cong (x_0,1)$ and so we have an alternative of looking of the mapping torus: Instead of saying it is $S^1 \vee S^1 \times I / \sim$, I can say that it is $S^1 \vee S^1$ wedged with another $S^1$ (to make it $S^1 \vee S^1 \vee S^1$) at $x_0$ slapped on with two $2$ - cells. The skeleton that we are slapping the cells onto is shown in the picture below:

enter image description here

From proposition $1.26$ of Hatcher, I get that the fundamental group of $T_f$ is

$$\pi_1(T_f,x_0) \cong \pi_1(S^1 \vee S^1 \vee S^1)/\langle acf_{\ast}(a^{-1})c^{-1}, bcf_{\ast}(b^{-1})c^{-1} \rangle.$$

My problem: For the second of these cases, namely when $X= S^1 \times S^1$ now, I have to think about slapping on cells onto the following skeleton:

enter image description here

However I can't seem to be able to get my head around how 2-cells or 3-cells are going to be slapped onto this one. First of all, I don't understand if I am going to need to slap on one $2$ - cell or two $2$ - cells. Heuristically from the case before, I will need to slap on two $2$ - cells. Furthermore don't I need to slap on at least one $3$ - cell? All these questions are based on intuition obtained from the previous example. Is my understanding of the case when $X = S^1 \times S^1$ correct?

Thanks.

Edit: I just had an idea on computing the mapping torus for $S^1 \times S^1$ based on my calculation of $S^1 \vee S^1$ before. Recall that $S^1 \times S^1$ can be built out of $S^1 \vee S^1$ by attaching one $2$ - cell to it. So to compute the mapping torus $T_f$ for $S^1 \times S^1$, can't I just say that first we throw in the two cell corresponding to the torus (which would be $aba^{-1}b^{-1}$ looking at the first picture above) and then throw in two $2$ - cells like in the first example?

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May be it is obvious but how are you attaching 2-cells to $S^{1} \vee S^{1} \vee S^{1}$? –  Shraddha Srivastava Jan 5 '13 at 18:30
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up vote 1 down vote accepted

Your idea in the edit is partially correct; you have to also add in a 3-cell. Basically, the mapping torus adds one cell of dimension $(k+1)$ for every original cell of dimension $k$. That's why in the first example you added another $S^1$ (which was a 1-cell) and two two-cells.

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