Lemma 2

Let X be a topological space, let U be an open subset of X and let $\sigma$ be a strong contraction of X onto X-U. Then $\sigma (U\times \mathbf{I})\subset \overline{U}$. In particular, $\sigma (U\times \{1\})$ is contained in the boundary of U.

Let $x\in U$. The set of real numbers $t\in \mathbf{I}$ such that $\sigma (x, t)\in U$ is open in $\mathbf{I}$ and contains 0; let $s$ be its least upper bound. We have $\sigma (x, s)\in \overline{U}$. If $s <1,\sigma (x, s)\notin U$, by definition of $s$; the same is true if $s= 1$ since $\sigma (x,1)\in X-U$. Consequently, $\sigma (x, s)\in$ Fr(U). By definition of a strong contraction, we then have $\sigma (x, s) =\sigma (\sigma (x, s),1) =\sigma (x,1)$; in particular, $\sigma (x,1)\in \overline{U}$. Consequently, $\sigma (x,1)$ belongs to the boundary $\overline{U}\cap (X-U)$ of U.

Soit $t\in \mathbf{I}$; si $\sigma (x, t)\notin U$, il vient encore $\sigma (x, t) =\sigma (\sigma (x, t),1) =$ $\sigma (x,1)$, donc $\sigma (x, t)\in \overline{U}$, d’où le lemme.