Sufficient Condition for Properness, Lee's Topological Manifolds Let $X$ be a topological space. We say that a sequence $(x_i)$ of points in $X$ diverges to infinity (in X) if for each compact subset $K\subseteq X$ there are at most finitely many indices $i$ such that $x_i \in K$.
Lemma 4.91 says: Suppose $X$ is a first countable Hausdorff space. A sequence in $X$ diverges to infinity if and only if it has no convergent subsequence.
Proposition 4.93 says: Let $F:X \to Y$ be a continuous map. If $X$ is a second countable Hausdorff space and $F$ takes sequences diverging to infinity in $X$ to sequences diverging to infinity in $Y$, then $F$ is proper.
I have a question about the proof. The proof is: let $K$ be a compact subset of $Y$, and we have to show that $F^{-1}(K)$ is compact. Because of the hypothesis on $X$, it suffices to show that $F^{-1}{(K)}$ is sequentially compact. Suppose on the contrary that $(x_i)$ is a sequence in $F^{-1}{(K)}$ with no convergent subsequence . By Lemma 4.91 the sequence $(x_i)$ diverges to infinity, so our assumption about $F$ implies that $(F(x_i))$ diverges to infinity. But this is impossible beacuse $F(x_i)$ lies in the compact set $K$ of $Y$ for all $i$.
My objection is: when I suppose that $F^{-1}(K)$ is not sequentially compact, then i can take a sequence $(x_i)$ in $F^{-1}(K)$ which has no subsequence converging to a point of $F^{-1}(K)$. So this sequence is diverging to infinity in the space $F^{-1}(K)$, and it could be that it does not diverge to infinity in the space $X$. Since the hypothesis is that $F$ takes sequences diverging to inifinity in $ X$ to sequences diverging to infinity in $Y$, then I think I can't say that $F((x_i))$ is a sequence diverging to infinity in $Y$.

Where is my mistake?


I believe $Y$ needs to be assumed Hausdorff for this to work.  In this case, $F^{-1}(K)$ is closed in $X$, and the proof runs without a problem.  
The following is a counterexample when we don't assume $Y$ is Hausdorff:  Let $X=\mathbb R$, let $Y$ be the line with two origins, which I denote $0_1$ and $0_2$, and let $F:X\to Y$ be the identity on $\mathbb R\setminus\{0\}$, and $F(0)=0_1$.  Then $F$ satisfies the hypotheses, but $K=\{0_b\}\cup(0,1]$ is compact in $Y$ while $F^{-1}(K)=(0,1]$ is not compact in $X$.  
