Let $f:[0,1]\rightarrow X$ be a continuous surjective function to a Hausdorff space $X$. Prove that $X$ has the following property:
For every $x\in X$ and every neighborhood $U$ of $x$, there exists a neighborhood $V$ of $x$ such that for all $a,b\in V$ there is a path from $a$ to $b$ contained in $U$.
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I have been going through this thinking of things that could potentially help. I realize that what we are trying to show is equivalent to showing that $X$ is locally path connected.
I was thinking about using that a space $X$ is locally path connected if and only if for every open set $U$ of $X$, each path component of $U$ is open in $X$.
I was also thinking of using that since $X$ is Hausdorff, we know that any closed subset is going to be compact, and then somehow intersecting the finite subcover with an open neighborhood of $x\in X$ to get a smaller neighborhood. (Edit: the comments have pointed out that this is false. I was thinking compact in Hausdorff is closed. That being said it made me think of how the image of a compact set under a continuous map is compact)
Edit: using that the image of [0,1] will be compact, can we still use the idea of a finite subcover to find a smaller neighborhood of $x$?
I was also thinking of how the image of a connected space under a continuous map is connected.
With all of these thoughts together, I was unsure of how to proceed. Any suggestions for how to proceed would be appreciated.
Best Answer
Lemma: $f$ is a closed map.
Proof: Let $C \subset [0, 1]$ be closed. Because $[0, 1]$ is compact, $C$ is compact. Then $f(C)$ is compact. Since $X$ is Hausdorff, this means that $f(C)$ is closed.
Now, let $x$ be any element of $X$ and $A$ be any open neighborhood of $x$. Choose an open interval $U_t$ around each element $t \in f^{-1}(x)$ such that $f(U_t) \subset A$. Then $U = \bigcup U_t$ is open. What do we know about $$X \setminus f\left([0, 1] \setminus U\right)?$$