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Let $f:N \to M$ be a continuous function between general topological spaces. Let's assume that each point $p \in N$ has an open neighborhood $U$ such that $f|_U$ is an injection. Then $f|_U$ is a continuous bijection onto its image. However, $f|_U$ may not be a homeomorphism, so we cannot expect $f$ to be a topological immersion in general. If $N$ and $M$ are topological manifolds (Hausdorff, second countable, locally Euclidean), is $f$ a topological immersion? I know that, if $M$ is Hausdorff and we can take compact $V \subset U$ for each $p$ then $f|_V$ must be a homeomorphism onto its image, and so $f$ is a topological immersion. I conjectured that it should be for manifolds but failed to argue.

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    What is your definition of a topological immersion between topological spaces? – 2017-02-14
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    The definition is a continuous map for which each point of the domain has a neighborhood that is mapped onto its image homeomorphically. – 2017-02-14
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    Oh, I see. Then all what you need is that $N$ is locally compact and $M$ is Hausdorff. Now, note that manifolds are locally compact. Incidentally, there is no consensus on what a "neighborhood" means. Some people require it to be open, some do not. – 2017-02-14
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    I'm sorry. My question had a bug. My neighborhood is an open set. I edited the question to be clear. In locally compact spaces, each point has a neighborhood $V$ whose closure is compact. But how can I sure that $\bar{V}$ is contained in the open set $U$ in my question? – 2017-02-14
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    The most common definition of "locally compact" is that every point has a basis consisting of compact neighborhoods: Even open neighborhood $U$ of $x$ contains a compact neighborhood of $x$. This is what you want to use. – 2017-02-14
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    @MoisheCohen Thank you. Now I understand it. – 2017-02-14
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    Isn't the definition proposed in the second comment precisely that of a local homeomorphism? – 2017-12-28

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