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I'm interested in continuous maps between topological spaces $f:X\to Y$ such that for any compact subset $L$ of $Y$ contained in $f(X)$, there is a compact subset $K$ of $X$ such that $L$ is contained in $f(K)$.

Proper maps satisfy this, but there are examples of continuous maps which don't, for example with discrete spaces, taking a non-stationary convergent sequence extended at infinity.

I would like to know if there are characterizations for those topological spaces which have enough compact subsets in the sense that: each real-valued continuous function satisfy this property.

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2 Answers 2

Every path-connected space $X$ satisfies this property. The image $f(X)$ is a connected subset of $\mathbb R$ and is hence a (possibly infinite, or trivial) interval. Every compact sub-interval $[y_0,y_1]$ is contained in the image of a compact path in $X$ connecting a point in $f^{-1}(y_0)$ to another point in $f^{-1}(y_1)$. Every compact subspace $L$ of $f(X)$ is contained in one such sub-interval, so you are done.

Note however that it is easy to construct an $X$ with two path-connected components which does not satisfies the property. For instance, take $X$ as two copies of $\mathbb R$ and define $f$ as constantly $\pi/2$ on one copy and as $f(x) = \arctan x$ on the other.

Added. If $Y = \mathbb C$ then this is no longer true. For instance, a (non-embedding) immersion $f:\mathbb R \to \mathbb C$ with compact image is a counterexample: simply take $L= f(\mathbb R)$.

immersion of a line in a plane which is not an embedding

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I don't understand why a continuous image of a path-connected set has to be an interval. $X={\bf R}^2$ is path-connected, the identity map $f:X\to X$ is continuous, and its image is $X$, which is not an interval. –  Gerry Myerson Oct 31 '10 at 10:30
    
Truly asks about real-valued maps, so $Y = \mathbb R$. –  Bruno Martelli Oct 31 '10 at 10:47
    
Ah, in the last clause of the question. I missed that, just saw that in the rest of the question things were more general. Thanks. –  Gerry Myerson Oct 31 '10 at 11:43
    
(I edited the answer to avoid ambiguities, thanks) –  Bruno Martelli Oct 31 '10 at 11:45
    
Thanks. So there's a large class of spaces whose real-valued continuous functions satisfy this. What if we require this for all complex-valued continuous function ? It looks much more restrictive. –  Truly Oct 31 '10 at 16:46

Such maps are called compact-covering maps, and are a somewhat well-known and well studied object. They came up naturally in many contexts, and if you look for that keyword in mathscinet, you will find many matches. As Bruno mentions, it is very easy to construct examples of maps in very ordinary settings that fail to be CC.

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