I am reading Allan Hatcher's Algebraic Topology and don't understand a step in one of its proofs. The theorem I am reading is theorem 2.10:

If two maps $f ;g :X \to Y$ are homotopic, then they induce the same homomorphism $f_\ast = g_\ast: H_n(X) \to H_n(Y)$.

In the proof, $\Delta^n$ denotes a $n$- simplex. The proof of the theorem starts with stating that the essential ingredient is a procedure for subdividing the product $\Delta^n \times [0,1]$, where the ground simplex has vertices $v_0, v_1, \ldots, v_n$ and the upper simplex has vertices $w_0, w_1, \ldots, w_n$ and the projection map $p: \Delta^n \times [0,1] \to \Delta^n$ maps the vertices $v_i, w_i$ onto the same points.

The proof then states:

The $n$-simplex $[v_0, \ldots, v_i, w_{i+1}, \ldots, w_n]$ is the graph of the linear function $\varphi_i: \Delta^n \to [0,1]$ defined in barycentric coordinates by $\varphi_i(t_0, \ldots, t_n) = t_{i+1} + \ldots + t_n$ sinec the vertices of this simplex $[v_0, \ldots, v_i, w_{i+1}, \ldots, w_n]$ are on the graph of $\varphi_i$ and the simplex projects homeomorhically onto $\Delta^n$ und the projection $\Delta^n \times [0,1] \to \Delta^n$.

I do not understand this part. Anyone who could give some more explanations?


A point in the $n$-simplex $[v_0,\dots,v_n]$ can be uniquely described by the sum $t_0 v_0 + t_1 v_1 + \cdots + t_n v_n$ where $0\leq t_i\leq 1$ for $i=1,\dots,n$. Thus each point in $\Delta^n$ can be uniquely described by the coefficients $(t_0,\dots, t_n)$ in the previous sum, called the barycentric coordinates of $\Delta^n$.

If $f:X\to Y$ is a function, then the graph of $f$ is the subset of $X\times Y$ defined by $$\operatorname{graph}(f) = \{(x,f(x)) \in X\times Y\}.$$ In our situation, the function $\phi_i$ is a function from $\Delta^n$ to the interval $[0,1]$, and so the $\operatorname{graph}(\phi_i) \subset \Delta^n\times [0,1]$. Of course $\Delta^n\times \{0\}$ and $\Delta^n\times \{1\}$ are $n$-simplices. Hatcher is using the vertices $[v_0,\dots,v_n]$ to describe the $n$-simplex $\Delta^n\times \{0\}$ and the vertices $[w_0,\dots, w_n]$ to describe the $n$-simplex $\Delta^n\times \{1\}$.

We want to further explain why the $n$-simplex $[v_0,\dots, v_i, w_{i+1},\dots, w_n]$ is the graph of $\phi_i$. Since $\phi_i(t_0,\dots,t_n) = t_{i+1}+\cdots+t_{n}$, a point on the graph of $\phi_i$ looks like $$((t_0,\dots, t_n),t_{i+1}+\cdots+t_{n})$$ in barycentric coordinates. The vertices $v_j$ have coordinates $(v_j,0)$ inside of $\Delta^n\times[0,1]$, and the vertices $w_j$ have coordinates $(v_j,1)$ inside $\Delta^n\times[0,1]$ (for each $j=0,\dots, n$).

Intuitively, it can help to consider pairs of indices $j$ and $k$, and look at what happens when all other barycentric coordinates are zero. Suppose that $0\leq j,k \leq i$ and $t_\ell=0$ for all other coordinates $\ell\neq j,k$. This subset of the graph of $\phi_i$ traces out a line between $(v_j,0)$ and $(v_k,0)$. Suppose that $i<j,k\leq n$ and $t_\ell=0$ for all other coordinates $\ell\neq j,k$. This subset of the graph of $\phi_i$ traces out a line between $(v_j,1)$ and $(v_k,1)$, that is, between $w_j$ and $w_k$. Suppose that $0\leq j \leq i$, $i<k\leq n$, and $t_\ell=0$ for all other coordinates $\ell\neq j,k$. This subset of the graph of $\phi_i$ traces out a line between $(v_j,0)$ and $(v_k,1)$, that is, between $v_j$ and $w_k$. Hopefully, this is enough intuition to see why the graph of $\phi_i$ is the $n$-simplex $[v_0,\dots, v_i,w_{i+1},\dots,w_n]$.

If we forget the last coordinate of each point in the graph of $\phi_i$, i.e. forget the coordinate corresponding to $[0,1]$, then we are projecting the point $((t_0v_0 + \cdots + t_nv_n),t_{i+1}+\cdots+t_n)$ in $\Delta^n\times[0,1]$ to the point $(t_0v_0+\cdots+t_nv_n)$ in $\Delta^n$. This projection map is a homeomorphism. It has the effect of flattening the $n$-simplex $\operatorname{graph}(\phi_i)=[v_0,\dots, v_i,w_{i+1},\dots,w_n]\subset \Delta^n\times[0,1]$ into the $n$-simplex $[v_0,\dots,v_n]\subset \Delta^n\cong \Delta^n\times\{0\}.$

Edit: I'll add a picture of what happens in the case the $n=2$ and we consider the graph of $\phi_1$. On the left is (my admittedly poor) depiction of $\Delta^2\times[0,1]$. On the right is the graph of $\phi_1$. It is a triangle (with its interior) whose vertices are $v_0$ and $v_1$ on the bottom and $w_2$ on the top. enter image description here

  • 1
    $\begingroup$ Very clear explanation! I confused the graph with the image, but even without this confusion of my part I would not have been able to understand this part so clearly! Thank you very much! $\endgroup$ – Student Dec 23 '17 at 20:25

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