Lemma 9

There exists a continuous, increasing and surjective mapping $u:\mathbf{I}\rightarrow \mathbf{I}$ and an injective path $c:\mathbf{I}\rightarrow X$ joining $a$ to $b$ such that $g=c\circ u$.

Let us first prove that the mapping $g$ is continuous. Let $t$ be a point of $\mathbf{I}$. If $t\in U,g$ is constant in a neighbourhood of $t$, hence continuous at $t$. Suppose that $t\notin U$ and let W be an open neighbourhood of $g(t)$ in X. Let V be an open interval in $\mathbf{I}$ containing $t$ such that $f(V)\subset W$; such an interval exists since $f$ is continuous at $t$. Let us prove that $g(V)\subset W$. Let $x\in V$; if $x\notin U$, we have $g(x) =f(x)$, hence $g(x)\in W$. Suppose then that $x\in U$ and let $]u, v[$ be the connected component of $x$ in U. Observe that $]u, v[$ does not contain $t$. Consequently, if $x < t$, we have $x < v\leqslant t$ whence $v\in V$ and $g(x) =g(v) =f(v)\in f(V)\subset W$. The same argument shows that $g(x)\in W$ if $x > t$. Thus, $g$ is continuous at $t$.

Let $u$ and $v$ be points of $\mathbf{I}$ such that $g(u) =g(v)$ and $u < v$. Let us prove that $]u, v[\subset U$. Put $U'= U\cup ]u, v[$; it is an open subset of $]0,1[$.

If $u$ belongs to U, let $u'$ denote the greatest lower bound in $\mathbf{I}$ of the connected component of $u$ in U; put $u'=u$ if $u$ does not belong to U. Analogously, if $v$ belongs to U, let $v'$ denote the least upper bound in $\mathbf{I}$ of the connected component of $v$ in U; put $v'=v$ if $v$ does not belong to U. Then, $]u', v'[$ is a connected component of $U'$ and one has $f(u') =g(u) =g(v) =f(v')$. Since the connected components of $U'$ distinct from $]u', v'[$ are connected components of U, the open set $U'$ is an element of $\mathscr{U}$. Since U is a maximal element of $\mathscr{U}$ for the relation of inclusion, one has $U'= U$ and $]u, v[$ is contained in U. This proves that $g$ is constant on the interval $[u, v]$. The fibres of $g$ are therefore intervals of $\mathbf{I}$, and these intervals are closed because $g$ is continuous.

Let R be the equivalence relation associated with $g$ and let $p$ be the canonical surjection of $\mathbf{I}$ onto $\mathbf{I}/R$. There exists a unique continuous mapping $g'$ of $\mathbf{I}/R$ into X such that $g=g'\circ p$; this mapping is injective. By the corollary, III, p. 276, of prop. 16, there exists a mapping $u$, increasing, continuous and surjective, such that R is the equivalence relation associated with $u$. Since the space $\mathbf{I}$ is compact and the space X is separated, the mapping $u$ is closed, hence strict (I, p. 18, example 2). It defines by passing to the quotient a homeomorphism $u'$ of $\mathbf{I}/R$ onto $\mathbf{I}$. Then, the mapping $g'\circ (u')^{-1}$ of $\mathbf{I}$ into X is an injective path of origin $a$ and of term $b$.