Bredon's Cone Construction Bredon has a slick proof based on the cone construction,  that if $X$ is a contractible topological space, then $H_i(X)=0$ for $i\neq 0$. My question has to do with a small detail:
Let $F:X\times I\to X$ be a homotopy from $1_X$ to a constant function $x_0\in X,\ $ set $t'=1-t_0,$ and for each singular simplex $\sigma:\Delta_{n-1}\to X$ define $D\sigma:\Delta_n\to X$ by 
$D\sigma(t_0e_0+\cdots +t_ne_n)=F\left ( \sigma \left ( \frac{t_1}{t'}e_0+\cdots \frac{t_n}{t'}e_{n-1} \right ),t_0 \right ),\ $
It is then straightforward to prove that $\partial D+D\partial=1-\epsilon,$ as desired. 
Now, $D\sigma,$ as it stands, is not defined at $t_0=1$, so I guess that Bredon is assuming without comment that $D\sigma(e_0)=x_0.$ 
My question is: how do we prove continuity of $F$ at $e_0?$
It is clear that $\left \{ \frac{t_1}{t'}e_0+\cdots \frac{t_n}{t'}e_{n-1} \right \}$ is bounded by $1$ as $t_i$ ranges over $0\leq t_i\le 1$ with $\Sigma t_i=1.$
Then, if $z_n\to e_0\in \Delta_n,\ $ this induces a sequence $\lambda_n\in I$ such that $\lambda_n\to 1$ and a sequence $w_n\in \Delta_{n-1}.$
Now, using compactness, we can find an $x\in X$ and subsequences $w_{n_k},\lambda_{n_k}$ such that $\sigma (w_{n_k})\to x$ and $\lambda_{n_k}\to 1.$ Continuity of $F$ now implies that $F(\sigma (w_{n_k}),\lambda_{n_k})\to F(x,1)=x_0.$
Of course, this is not enough, since I was forced to pass to a subsequence. 
So, my questions are: is the case $t_0=1$ so trivial that Bredon left it out of his definition? How do we prove continuity of $D\sigma$ on all of $\Delta_n?$
 A: First of all, what Bredon presents here is of course the standard proof.
Your analysis is absolutely right, also regarding the gap in your argument. It is an easy to close one: Instead of starting with the sequence $z_n$, start with an arbitrary subsequence. This way you show that every subsequence of $D\sigma(z_n)$ has a subsequence converging to $x_0$, and this suffices to show that $D\sigma(z_n)$ converges to $x_0$.
I would have formulated the same idea differently. The idea is to first define a map from the cone over $\Delta_{n-1}$ and then to identify that cone with $\Delta_n$. We first consider $g:=F\circ(\sigma\times\mathop{id}_I)\colon \Delta_{n-1}\times I\to X$. Since $g[\Delta_{n-1}\times \{1\}]\subset F[X\times\{1\}] = \{x_0\}$, this induces a continuous function $\bar g\colon C\Delta_{n-1}:=(\Delta_{n-1}\times I)/(\Delta_{n-1}\times \{1\})\to X$.
Now for the identification of $C\Delta_{n-1}$ with $\Delta_n$, consider the map $\Delta_{n-1}\times I\to \Delta_n$, $\Phi\colon\left(t_1e_0+\cdots t_n e_{n-1},t_0 \right)\mapsto t_0e_0+(1-t_0)t_1+\cdots +(1-t_0)t_ne_n$. Since $\Phi(x, t)=\Phi(x', t')$ iff $t=t'=1$ or ($t=t'$ and $x=x'$), this map induces a continuous bijection $\phi\colon C\Delta_{n-1}\to\Delta_n$. Because $C\Delta_{n-1}$ is quasi-compact and $\Delta_n$ hausdorff, the map $\phi$ is a homeomorphism. This is the technical analogue of your argument.
Now $g\circ \phi^{-1}$ is the function whose continuity you wanted to show.
Thanks for your comments which made me get the second part of the answer into shape.
A: Here's the answer suggested in my comments above.
We want to show $D\sigma$ is continuous at $t_0=1$. So pick an open neighborhood $x_0\in V$; we need to find an open neighborhood of $(1,0,0,\ldots,0)$ which maps into $V$.
Because $F$ is continuous, and $F(\cdot, 1)=x_0$, for every $x\in\sigma(\Delta_{n-1})$, we can find an open neighborhood $x\in U_x$ and a real number $r_x$ such that
$$ F\left(U_x, (r_x,1]\right)\subset V$$
Since $\sigma(\Delta_{n-1})$ is compact, we can find finitely many $U_i$ covering it, and if we let $r$ be the maximum of the corresponding $r_i$, we see that 
$$ F\left(\sigma(\Delta_{n-1}), (r,1]\right)\subset F\left(\cup U_i, (r,1]\right)\subset V$$
So if we have $t_0>r$, we then see that 
$$D\sigma(t_0e_0+\cdots +t_ne_n)\subset V$$
Thus the open subset of $(1,0,0,\ldots,0)$ given by $t_0>r$ maps into $V$, and this establishes continuity of $D\sigma$ at $t_0=1$.
