Continuity of taking collapse maps Let $U$ and $V$ be open subsets of $\mathbb R^n$ and let $\mathrm{OEmb}(U,V)$ denote the space of open embeddings of $U$ into $V$ with the compact-opent topology. Let $\bar{U},\bar{V}$ denote their one-point compactifications and let $\mathrm{map}(\bar{V},\bar{U})$ denote the space of maps between them again with the compact-open topology. There is a (set-theoretic) map
$$\mathrm{OEmb}(U,V)\longrightarrow \mathrm{map}(\bar{V},\bar{U})$$
that sends an embedding $e$ to the map $\phi(e)$ given by $\phi(e)(y) = e^{-1}(y)$ if this value exists and $\phi(e)(y)=\infty$ otherwise.
Fact: This map is continuous. More generally, it is also continuous if $U$ and $V$ are Hausdorff locally compact locally connected topological spaces. 
What reference should I cite for these facts? 
 A: I haven't found an explicit reference in the literature, but as Johannes Ebert pointed to me, in the second part of Theorem 4 of
R. Arens, Topologies for Homeomorphism Groups, American Journal of Mathematics,  Vol. 68, No. 4 (Oct., 1946), pp. 593-610
it is proven that the inverse mapping $\mathrm{Homeo}(X)\to \mathrm{Homeo}(X)$ that sends a homeomorphism to its inverse is continuous with respect to the compact-open topology, provided that $X$ is Hausdorff, locally compact and locally connected. The same proof works as well to prove my question (under the assumptions that $X$ and $Y$ are Hausdorff, locally compact, and $Y$ is locally connected). Therefore I think that Arens article is a good reference for this fact (together with an indication saying that the proof has to be mildly adpated). This is how the adaptation would go:
Theorem: If $X$ and $Y$ are Hausdorff locally compact spaces and $Y$ is locally connected, then the collapsing map
$$\phi\colon \mathrm{OEmb}(X,Y)\longrightarrow \mathrm{map}(\overline{Y},\overline{X})$$
is continuous with the compact-open topologies.
Proof: Take an embedding $e$ and consider a subbasic neighbourhood $(K,U)$ of $\phi(e)$, where $K\subset \overline{Y}$ is compact and $U\subset \overline{X}$ is open. Let us denote by $U^c$ the complement of $U$ in $X$ (not in $\overline{X}$). An embedding $f$ belongs to this neighbourhood if $\phi(e)(K)\subset U$, and this holds if and only if $e(U^c)\subset K^c$. Therefore $e(U^c)\subset K^c$ and if any other embedding $f$ satisfies that $f(U^c)\subset K^c$, then it holds that $\phi(f)\in (K,U)$.
We distinguish two cases:
\underline{If $\infty\in U$}, then $U^c$ is compact and we are done ($(U^c,K^c)$ is an open neighbourhood of $e$ that gets mapped into the neighbourhood $(K,U)$ of $\phi(e)$.
\underline{If $\infty\notin U$}, then $K\subset e(X)$, and therefore $e^{-1}(K)$ is compact. This is the situation that Arens handles. We use that $X$ is Hausdorff locally compact to construct a pair of open neighbourhoods $V,W$ of $K$ with compact closure of $e$ such that 
$$e^{-1}(K)\in V\subset \overline{V}\subset W\subset \overline{W}\subset U.$$
Then $N:=\overline{W}\setminus V$ is compact, and if $f\in (N,K^c)$, then $\phi(f)(K)\subset N^c$. Observe that $N^c$ is the disjoint union of $V$ and $\overline{W}^c$. 
Now we use that $X$ is locally connected to assume, without loss of generality that $K$ is connected (this is detailed in Arens' paper). Therefore, if $f\in (N,K^c)$, then either $\phi(f)(K)\subset V\subset U$ (this is what we want) or $\phi(f)(K)\subset \overline{W}^c$. In order to rule out the second case, we make the additional assumption (again without loss of generality) that $K$ has non-void interior, then we take a point $p$ in the interior. If
$f\in (N,K^c\cap e(W))\cap (\{e^{-1}(p)\},\mathring{K})$,
then $f(e^{-1}(p))\subset K$, but then $e^{-1}(p)\in \phi(f)(K)\cap V$, which is non-empty, and therefore we have ruled out the second case and $f\in (K,V)\subset (K,U)$.
