Lemma 2

Let us retain the notation and hypotheses of the proposition. Let T be a set endowed with an operation of the groupoid Coeg($f$) relative to a mapping $q: T\rightarrow Y$. Then there exists a unique topology on T for which $q$ makes T a covering of Y, such that for every class of paths with origin $c$ in X and every point $t$ of T, one has $t\cdot \gamma (c) =t\cdot f_*(c)$. In particular, the operation of Coeg($f$) on T factors through an operation of the groupoid $\varpi (Y)$.

The uniqueness of such a topology on T follows from Lemma 1 (III, p. 312); let us prove its existence. Endow the set $X\times_YT$ with a law of operation of $\varpi (X)$ by putting $(x, t)\cdot u= (x\cdot u, t\cdot \gamma (u))$ for every $(x, t)\in X\times_YT$ and every arrow $u$ with origin $x$ in $\varpi (X)$. Since the space X is simply connected, this law of operation of $\varpi (X)$ is without local monodromy (III, p. 313, Remark). There therefore exists on the set $X\times_YT$ a unique topology which makes the X-space $(X\times_YT$, pr$_1)$ a covering of X and such that the canonical operation of $\varpi (X)$ in this covering coincides with the given operation (III, p. 313, Prop. 3). Let us denote this covering by $(Z, p)$. Let us prove that the mapping $\tau : ((x_1, t), x_2)\mapsto (x_2, t)$ of $Z\times_YX$ into Z is a descent datum on Z relative to the mapping $f$. It is enough to verify that it is continuous, the other conditions of Definition 1 of IV, p. 382, being evident.

The mapping $\tau$ is a lifting of the mapping pr$_2: Z\times_YX\rightarrow X$ to the covering Z of X. Since the space $X\times_YX$ is locally arcwise connected, the space $Z\times_YX$, which is a covering of it, is also locally arcwise connected. By the Corollary (III, p. 269), in order to prove that the mapping $\tau$ is continuous, it is enough to prove that it is arcwise continuous.

Let therefore $\widetilde{c}= ((c, g), c')$ be a path in $Z\times_YX$ and let us prove that the mapping $\tau \circ \widetilde{c}:t\mapsto (c'(t), g(t))$ of $\mathbf{I}$ into Z is continuous. For every $s\in [0,1]$, let us denote by $c_s$ and $c'_s$ the paths in X defined by $t\mapsto c(st)$ and $t\mapsto c'(st)$. For every $s\in [0,1]$, one thus has $c(s) =c(0)\cdot [c_s]$, $c'(s) =c'(0)\cdot [c'_s]$ and $(c, g)(s) = (c(0), g(0))\cdot [c_s]$, where $[u]$ denotes the strict homotopy class of a path $u($III, p. 304, Remark). The mapping $t\mapsto (c_s(t), c'_s(t))$ is a path in $X\times_YX$; by definition of the coequalizer Coeg($f$), one therefore has $\gamma ([c_s]) =\gamma ([c'_s])$ and $g(s) =g(0)\cdot \gamma ([c_s]) =$ $g(0)\cdot \gamma ([c'_s])$. By definition of the operation of $\varpi (X)$ on Z, one therefore has

This shows that $\tau \circ \widetilde{c}$ is a continuous lifting to Z of the path $c'($loc. cit.). Consequently, the mapping $\tau$ is arcwise continuous, hence continuous; it is a descent datum relative to $f$ on the X-space Z.

Let $R_{\tau}$ denote the equivalence relation defined by the descent datum $\tau$. Since the mapping $f$ is surjective, the mapping pr$_2: Z\rightarrow T$ induces, by passing to the quotient, a bijection of $Z/R_{\tau}$ onto T. Let us endow T with the topology deduced from that of $Z/R_{\tau}$ by transport of structure, so that $(T, q)$ is a Y-space. By hypothesis, T is therefore a covering of Y and the diagram

Z $^{pr_2}$ T

$pq$

X $^f$ Y

is a cartesian square.

Let $(x, t)$ be a point of Z, $c$ a path with origin $x$ in X, and let $\widetilde{c}$ be the continuous lifting of $c$ to Z with origin $(x, t)$. The path pr$_2\circ \widetilde{c}$ is the lifting with origin $t$ of the path $f\circ c$ in Y, so that $t\cdot \gamma ([c]) =$ $t\cdot [f\circ c] =t\cdot f_*([c])$. This completes the proof of the lemma.

Let us now prove the proposition. Let $u$ and $v$ be two arrows of Coeg($f$) whose images in $\varpi (Y)$ are equal. Since the groupoid morphism $\varpi '(f)$ is the identity on the vertex sets, the arrows $u$ and $v$ have the same origin and the same target. Let us denote by $y$ the origin of $u$; let T be the set of arrows of Coeg($f$) with origin $y$, and let $q: T\rightarrow Y$ be the restriction to T of the target mapping. The groupoid Coeg($f$) operates by composition on the right on the set T, relative to $q$. By lemma 2, the actions of $u$ and $v$ on T are identical. Hence $u=$ $e_y\cdot u=e_y\cdot v=v$. This proves that the groupoid morphism $\varpi '(f)$ is injective.