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No. There are two (discontinuous) surjective maps $f,g:S^1\to S^1$ whose graphs are connected but the graph of $g\circ f$ (as well as its closure) is not. The map $f$ is defined as follows, using th …

This is possible even on the real line. There is a strictly increasing continuous function $f:[0,1]\to\mathbb R$ whose derivative is zero almost everywhere. It is a suitable sum of a series of Cantor …

Consider $X=Y=S^1$. Let $f:X\to Y$ be a 2-fold covering and $g:X\to\mathbb R^2$ the standard embedding (whose image is a unit circle). Assume $\epsilon<1$, then there is no map $h$ with the desired pr …

Yes. Let $A$ be the set in question. We may assume that $0\notin A$ and moreover that $A$ is outside the unit ball centered at the origin. Since $A$ is closed (in a complete space) and not compact, i …

Enumerate all rational points outside your set. Then cover these points by open balls by induction as follows: the next ball is centered at the first rational point not covered so far, its radius is s …

For $A=[0,1]$, let $B$ be the 2-torus with one hole and $C$ be the 2-disc with two holes. The products $B\times[0,1]$ and $C\times[0,1]$ can be realized in $\mathbb R^3$: the former as a thickening o …

Here is a simple proof for the Hausdorff dimension. Consider the orthogonal projection to $\mathbb R^{d-1}$. Since $A$ is bounded, the projection of the boundary contains the projection of $A$. The la …

This is not a reference but a short direct proof. Let $\bar X$ be the completion of $X$. Define $f:X\to\mathbb R$ by $f(x)=dist(x,\bar X\setminus X)$. Obviously $f$ is continuous, and the local compl …

No, it is not even path-connected in general, already for $n=1$. Consider the folding map $f:[0,1]\to[0,1]$, namely $f(t)=2t$ for $t\le 1/2$ and $f(t)=2(1-t)$ for $t\ge 1/2$. There is no path connect …

No for general topological spaces, yes for metrizable ones (and I believe the argument can be generalized to all normal spaces). Bad example: $X=\{a,b,c\}$ with open sets $\emptyset$, $X$, $\{a\}$, $ …

No. Let $B=S^1$, $F=\mathbb Z$, $E=S^1\times\mathbb Z$, and let $p$ be a two-fold covering on each component: $p(z,n)=z^2$ where $S^1$ is regarded as the unit circle in $\mathbb C$ (for the purposes o …

Here is a counter-example with $G$ homeomorphic to $\mathbb R^2$. Let $f:\mathbb R\to\mathbb R$ be a discontinous additive homomorphism (constructed using a Hamel basis of $\mathbb R$ over $\mathbb Q$ …

The comments explain how to prove the fact. If you want to put a formal wrapping around it, consider the strong (Whitney) $C^\infty$ topology on the space of maps $W\to M$. The strong $C^0$ topology …

I can partially answer the second question. If $X$ is a compact Hausdorff space whose topology has a countable base at every point [Edit: $\omega_1$ has this property but not compact], then there are …

Here is a formalization of André Henriques' answer to the Hausdorff dimension variant of the question. Let $K=\{0,1\}^\infty$ be the standard Cantor set. Define a map $f:K\to[0,1]$ as follows: for a …