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Martin Sleziak
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There is, probably, no hope for a similar characterization since for any topological space Y$Y$ and any X$X$, which is line-filling in your terminology, if f: X \to [0,1]$f: X \to [0,1]$ is continuous and surjective then h: X \cap Y \to [0,1]$h: X \sqcup Y \to [0,1]$, where X \cap Y$X \sqcup Y$ is a sum of spaces (=disjoint union) X$X$ and Y$Y$, given by h|_X = f$h|_X = f$ and h(Y) = 0$h(Y) = 0$ is also surjective and continuous. So your spaces maybe as bizzarebizarre as you want.

Update: Alejandro, thank you for your comment. I don't think though, that connectedness (or path-connectedness) would make things better. Again, for any connected (path-connected) Y and any line-filling space X, f: X \to [0$X$,1] $f: X \to [0,1]$ take x \in X$x \in X$ and y \in Y$y \in Y$ and glue X$X$ and Y$Y$ at the point (x,y)$(x,y)$ (that is consider equivalence relation with only one non-trivial equivalence x \equiv y$x \equiv y$ and take the factor space). Denote the result by Z$Z$. Then one extends f$f$ to Z$Z$ by defining f(Y) = f(x)$f(Y) = f(x)$. I believe Z$Z$ is then connected (if X$X$ and Y$Y$ were) and again pretty random.

There is, probably, no hope for a similar characterization since for any topological space Y and any X, which is line-filling in your terminology, if f: X \to [0,1] is continuous and surjective then h: X \cap Y \to [0,1], where X \cap Y is a sum of spaces (=disjoint union) X and Y, given by h|_X = f and h(Y) = 0 is also surjective and continuous. So your spaces maybe as bizzare as you want.

Update: Alejandro, thank you for your comment. I don't think though, that connectedness (or path-connectedness) would make things better. Again, for any connected (path-connected) Y and any line-filling space X, f: X \to [0,1] take x \in X and y \in Y and glue X and Y at the point (x,y) (that is consider equivalence relation with only one non-trivial equivalence x \equiv y and take the factor space). Denote the result by Z. Then one extends f to Z by defining f(Y) = f(x). I believe Z is then connected (if X and Y were) and again pretty random.

There is, probably, no hope for a similar characterization since for any topological space $Y$ and any $X$, which is line-filling in your terminology, if $f: X \to [0,1]$ is continuous and surjective then $h: X \sqcup Y \to [0,1]$, where $X \sqcup Y$ is a sum of spaces (=disjoint union) $X$ and $Y$, given by $h|_X = f$ and $h(Y) = 0$ is also surjective and continuous. So your spaces maybe as bizarre as you want.

Update: Alejandro, thank you for your comment. I don't think though, that connectedness (or path-connectedness) would make things better. Again, for any connected (path-connected) Y and any line-filling space $X$, $f: X \to [0,1]$ take $x \in X$ and $y \in Y$ and glue $X$ and $Y$ at the point $(x,y)$ (that is consider equivalence relation with only one non-trivial equivalence $x \equiv y$ and take the factor space). Denote the result by $Z$. Then one extends $f$ to $Z$ by defining $f(Y) = f(x)$. I believe $Z$ is then connected (if $X$ and $Y$ were) and again pretty random.

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There is, probably, no hope for a similar characterization since for any topological space Y and any X, which is line-filling in your terminology, if f: X \to [0,1] is continuous and surjective then h: X \cap Y \to [0,1], where X \cap Y is a sum of spaces (=disjoint union) X and Y, given by h|_X = f and h(Y) = 0 is also surjective and continuous. So your spaces maybe as bizzare as you want.

Update: Alejandro, thank you for your comment. I don't think though, that connectedness (or path-connectedness) would make things better. Again, for any connected (path-connected) Y and any line-filling space X, f: X \to [0,1] take x \in X and y \in Y and glue X and Y at the point (x,y) (that is consider equivalence relation with only one non-trivial equivalence x \equiv y and take the factor space). Denote the result by Z. Then one extends f to Z by defining f(Y) = f(x). I believe Z is then connected (if X and Y were) and again pretty random.

There is, probably, no hope for a similar characterization since for any topological space Y and any X, which is line-filling in your terminology, if f: X \to [0,1] is continuous and surjective then h: X \cap Y \to [0,1], where X \cap Y is a sum of spaces (=disjoint union) X and Y, given by h|_X = f and h(Y) = 0 is also surjective and continuous. So your spaces maybe as bizzare as you want.

There is, probably, no hope for a similar characterization since for any topological space Y and any X, which is line-filling in your terminology, if f: X \to [0,1] is continuous and surjective then h: X \cap Y \to [0,1], where X \cap Y is a sum of spaces (=disjoint union) X and Y, given by h|_X = f and h(Y) = 0 is also surjective and continuous. So your spaces maybe as bizzare as you want.

Update: Alejandro, thank you for your comment. I don't think though, that connectedness (or path-connectedness) would make things better. Again, for any connected (path-connected) Y and any line-filling space X, f: X \to [0,1] take x \in X and y \in Y and glue X and Y at the point (x,y) (that is consider equivalence relation with only one non-trivial equivalence x \equiv y and take the factor space). Denote the result by Z. Then one extends f to Z by defining f(Y) = f(x). I believe Z is then connected (if X and Y were) and again pretty random.

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There is, probably, no hope for a similar characterization since for any topological space Y and any X, which is line-filling in your terminology, if f: X \to [0,1] is continuous and surjective then h: X \cap Y \to [0,1], where X \cap Y is a sum of spaces (=disjoint union) X and Y, given by h|_X = f and h(Y) = 0 is also surjective and continuous. So your spaces maybe as bizzare as you want.