g(x)$.

Let $C \colon {{\mathcal{V}_{}}} \to {{\mathcal{U}_{}}}$ be the map defined by $C(V_U) = U$. It is a carrier and both $f$ and $g$ are carried by $C$, directly from the definition of ${{\mathcal{V}_{}}}$. Let $D \colon {{\mathcal{W}_{}}} \to {{\mathcal{V}_{}}}$ be a map such that for each $W \in {{\mathcal{W}_{}}}$ we have $W \subset D(W)$. The map $D$ exists because ${{\mathcal{W}_{}}}$ refines ${{\mathcal{V}_{}}}$. Let $E \colon {{\mathcal{F}_{}}} \to {{\mathcal{W}_{}}}$ be a map defined by $D(W \times [0,1]) = W$. The composition $C \circ D \circ E$ is a carrier and $H_0$ is carried by it. Observe that ${{\mathcal{U}_{}}}$ is an $AE(X \times [0,1])$-cover of $Y$ (see Definition \[def:AE(C)\]). By the Carrier Theorem, $H_0$ can be extended to a map $H \colon X \times [0,1]$ that is carried by $C \circ D \circ E$. Clearly, $H$ is a homotopy between $f$ and $g$. For each path $\{ x \} \times [0, 1] \subset X \times [0,1]$ there exists element $W \times [0, 1] \in {{\mathcal{F}_{}}}$ such that $\{ x \} \times [0, 1] \subset W \times [0, 1]$, as ${{\mathcal{W}_{}}}$ is a cover of $X$. Since $H$ is carried into ${{\mathcal{U}_{}}}$, the whole path lies in an element of ${{\mathcal{U}_{}}}$. Therefore 