Is a continuous image of $S^1$ to Hausdorff space locally connected?

Question

Is a continuous image of $S^1$ to Hausdorff space locally connected? How do you prove this?

Answer 1

The continuous surjective map $f:S^1\to X$ is closed because $S^1$ is compact and $X$ is Hausdorff.
Hence $f$ is a quotient map, like all surjective closed continuous maps.
The space $X$ being a quotient of a locally connected topological space (namely $S^1$) is also locally connected: the proof is not difficult and can be found in Bourbaki's General Topology, Chapter 1, §11, Proposition 12.

Answer 2

The image is clearly weakly locally connected at every point. Therefore it is locally connected. See Wikipedia article on local connectivity.

Edit: Here are some details:

Let $K$ be a compact metric space (e.g., the circle). Let $Y$ be a Hausdorff space and $f: X\to Y$ a continuous map. For a subset $C\subset X$ let $B(C,\epsilon)$ denote the $\epsilon$-neighborhood of $C$ in $K$, i.e., union $$ \bigcup_{c\in C} B(x, \epsilon). $$

Next lemma follows from continuity of $f$ and compactness of $K$:

Lemma. Let $C=f^{-1}(y)$ for some $y\in Y$ and $V\subset Y$ be a neighborhood of $y$. Then there exists $\epsilon>0$ such that $f(B(C,\epsilon))\subset V$.

Definition. The space $Y$ is called weakly locally connected at $y\in Y$ if for every neighborhood $V$ of $y$, there exists a neighborhood $W\subset V$ of $y$, such that every two points $z, w\in W$ belong to the same connected component of $V$. The space $Y$ is weakly locally connected if it is weakly locally connected at every point. The space $Y$ is called locally connected if for every $y\in Y$ if there exists a basis of neighborhoods of $y$ consisting only of connected subsets.

It is known that $Y$ is weakly locally connected if and only if it is locally connected.

Lemma. Suppose that $Y$ is Hausdorff, $K=S^1$ and $f: K\to Y$ is a surjective continuous map. Then $Y$ is weakly locally connected.

Proof. Pick $y\in Y$ and let $C=f^{-1}(y)$ and let $V$ be a neighborhood of $y$ in $Y$. By lemma, there exists $\epsilon>0$ such that $f(B(C, \epsilon))\subset U$. For each $x\in B= B(C, \epsilon)$, pick a point $c\in C$ such that $d(x,c)<\epsilon$. Let $xc$ denote the arc of $S^1$ connecting $x$ and $c$ and having length $<\epsilon$. Then $f(xc)\subset U$ and $f(xc)$ is connected. Therefore, $y$ and $f(x)$ belong to the same (path) connected component of $U$.

Lastly, observe that, by continuity of $f$, Hausdorff property of $Y$ and compactness of $K$, the interior $W$ of $f(B)$ contains the point $y$. Therefore, every two points $v, w\in W$, belong to the same (path) connected component of $U$. Thus, $Y$ is weakly locally connected. qed

With small modifications, the same proof works if $K$ is an arbitrary compact locally connected metrizable space. (Maybe even metrizable is not needed, I did not check.)