Show that image of simplex contains its barycenter Let $T^n$ be a standard $n$-simplex and $f_\varepsilon \colon T^n \to T^n$ be a family of continuous mappings, $\varepsilon > 0$. Furthermore $f_\varepsilon \to \mathop{\mathrm{id}}\nolimits_{T^n}$ in $C(T^n,T^n)$ when $\varepsilon \to +0$. How to show that for small $\varepsilon$ there is an $x(\varepsilon) \in  T^n$ such that $f_\varepsilon(x(\varepsilon)) = x^\ast$ where $x^\ast$ is the barycenter of $T^n$?
 A: I have taken the liberty to interpret the converge as a homotopy from $f_1$ to $f_0=$ Id$_X$. Let $X=T^n$ and assume that the barycenter $b$ is not in the image $f_t(T^n)$ for positive $t$. The boundary of $T^n$ is homeomorphic to $S^{n-1}$. We can restrict $f_t$ to this boundary to get a map $g_t:S^{n-1}\to T^n-{b}$ since $g_0=$Id and $g_t(S^{n-1})\not\ni b$. Now $g_0$ is the embedding of $S^{n-1}$ as the boundary and induces the identity on $\tilde H_{n-1}(S^{n-1})\approx\Bbb Z$ since $S^{n-1}\xrightarrow{g_0}T^n-\{b\}\ \xrightarrow{r}S^{n-1}$ (where $r$ is the deformation retraction onto the boundary) is the identity and induces the identity on homology $\tilde H_{n-1}(S^{n-1})\xrightarrow{g_{t*}}\tilde H_{n-1}(T^n-\{b\})\xrightarrow{r_*}\tilde H_{n-1}(S^{n-1})$. But $g_t$ is null-homotopic for $t>0$ since it factors as $S^{n-1}\hookrightarrow T^n\xrightarrow{f_t}T^n-{b}$, so it induces the zero map on homology.
A: More generally: if $f:T^n\to T^n$ is a continuous map and $|f(x)-x|<\operatorname{dist}(x^*,\partial T^n)$ for all $x\in \partial T^n$, then $x^*\in f(T^n)$. 
This is basically Rouché's theorem. The assumption implies that for $0\le t\le 1$ the map $f_t(x)=x-tf(x)$ does not attain the value $x^*$ on $\partial T^n$. The homotopy invariance of degree implies that $\deg(f,T^n,x^*)=\deg(\mathrm{id},T^n,x^*)=1$.
