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Let $X$ and $Y$ be topological spaces and $f:X\rightarrow Y$ and $g:Y\rightarrow X$ be surjective continuous maps. Is it necessarily true that $X$ and $Y$ are homeomorphic? I feel like the answer to this question is no, but I haven't been able to come up with any counter example, so I decided to ask here.

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Thanks for the answers everyone! –  Seth Jul 11 '12 at 14:19
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5 Answers

up vote 32 down vote accepted

The circle $S^1$ surjects onto the interval $I = [-1,1]$ by projection in (say) the $x$-coordinate, while the interval $I$ surjects onto the circle by wrapping around, say $f(x) = (\cos \pi x, \sin \pi x)$.

Added: Why is the circle $S^1$ not homeomorphic to the interval $I = [-1,1]$ ? The usual proof looks at cut points, i.e. a point $x$ whose removal from a topological space $X$ results in a disconnected space $X\backslash\{x\}$. Since this is a purely topological property, two homeomorphic spaces will have an equal number of cut points.

Note that $S^1$ has no cut points; removal of any single point from the circle leaves a connected open arc. However a closed interval $I$ has infinitely many cut points because removing any point except one of the two endpoints disconnects it into two disjoint subintervals.

The same observation serves to show the spaces in Karolis Juodelė's answer are not homeomorphic: $[0,1]$ has cut points and $[0,1]^2$ does not.

See Seth Baldwin's comment below for an alternative idea, something that will not disconnect the interval $I$ that does disconnect the circle!

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I like this one - no need for exotic Peano curves. –  akkkk Jul 11 '12 at 13:51
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Thanks, I should have perhaps have also pointed out why the circle is not homeomorphic to the interval (hint: no homology groups needed!). It seems to be a fairly common exercise... –  hardmath Jul 11 '12 at 13:55
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You can find two points on the closed interval to remove such that it stays connected, while this is impossible for the circle. –  Seth Jul 11 '12 at 14:24
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@SethBaldwin: That's neat, in that I was thinking about roughly the reverse, being able to disconnect the interval by removing one point (which we cannot do for the circle). –  hardmath Jul 11 '12 at 15:47
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@Theorem: I'll be happy to add a note explaining that, but see the above comment by Seth Baldwin and my reply. –  hardmath Jul 11 '12 at 17:08
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There is a continuous surjective map from $[0, 1]$ to $[0, 1]^2$ - the Peano curve. There is also a map from $[0, 1]^2$ to $[0, 1]$ - $f(x, y) = x$. However the two spaces are not homoemorphic.

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More generally: any two connected, locally connected, compact, second-countable spaces have your property. (Hahn-Mazurkiewicz Theorem)

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Strictly speaking, I must add that they have at least two points. –  GEdgar Jul 12 '12 at 19:35
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Others have answered, but maybe these comments will also be useful for this thread. A stronger equivalence (replace “surjective” with “bijective” in both places), which is still strictly weaker than being homeomorphic, was studied by several people in the early days of topology (Banach, Kuratowski, Hausdorff, Sierpinski, etc. in the 1920's). I believe this stronger equivalence originates from Frechet (1910), who called it type de dimensions. There is a lot about this relation in Sierpinski's General Topology (where it's called dimensional type; see pp. 130-133, 137, 141, 142, 144, 145, 163, 165) and in Kuratowski's Topology (where it's called topological rank).

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Just to be different, here’s an example that isn’t related to the Hahn-Mazurkievicz theorem. The example is originally due (with a different purpose) to K. Sundaresan, Banach spaces with Banach-Stone property, Studies in Topology (N.M. Stavrakas & K.R. Allen, eds.), Academic Press, New York, 1975, pp. 573-580; the argumentation is mine, On an example of Sundaresan, Top. Procs. 5 (1980), pp. 185-6. The surjections are $1$-$1$ save at a single point each, where they are $2$-$1$.

Let $X=\omega^*\cup(\omega\times 2)$, where $\omega^*=\beta\omega\setminus\omega$, and $2$ is the discrete two-point space, let $\pi:X\to\beta\omega$ be the obvious projection, and endow $X$ with the coarsest topology making $\pi$ continuous and each point of $\omega\times 2$ isolated. Let $N=\omega\times 2$, for $n\in\omega$ let $P_n=\{n\}\times 2$, and let $\mathscr{P}=\{P_n:n\in\omega\}$. A function $f:X\to X$ preserves pairs if $f[P_n]\in\mathscr{P}$ for all but finitely many $n\in\omega$.

Lemma. Let $f:X\to X$ be an embedding; then $f$ preserves pairs.

Proof. Suppose that $f$ does not preserve pairs. Since $f$ is injective, an easy recursion suffices to produce an infinite $M\subseteq\omega$ such that $(\pi\circ f)\upharpoonright\bigcup\{P_n:n\in M\}$ is injective. Let $M_i=M\times\{i\}$ for $i\in 2$. Then $$\left(\operatorname{cl}_XM_i\right)\setminus N=\left(\operatorname{cl}_{\beta\omega}M\right)\setminus\omega\ne\varnothing$$ for $i\in 2$, so $$\left(\operatorname{cl}_Xf[M_0]\right)\setminus N=\left(\operatorname{cl}_Xf[M_1]\right)\setminus N\ne\varnothing\;.$$ But $$\left(\operatorname{cl}_Xf[M_i]\right)\setminus N=\left(\operatorname{cl}_{\beta\omega}f[M_i]\right)\setminus\omega$$ for $i\in 2$, $\pi\big[f[M_0]\big]\cap\pi\big[f[M_1]\big]=\varnothing$, and disjoint subsets of $\omega$ have disjoint closures in $\beta\omega$, so $\operatorname{cl}_Xf[M_0]\cap\operatorname{cl}_Xf[M_1]=\varnothing$; this is the desired contradiction. $\dashv$

Now let $p$ be any point not in $X$, and let $Y=X\cup\{p\}$, adding $p$ to $X$ as an isolated point.

Proposition. $Y$ is not homeormorphic to $X$.

Proof. Suppose that $h:Y\to X$ is a homeomorphism; it follows from the lemma that $h\upharpoonright X$ preserves pairs. Let $$A=\bigcup\Big\{P_n\in\mathscr{P}:h[P_n]\in\mathscr{P}\Big\}\cup\omega^*\;.$$ Then $\big|X\setminus h[A]\big|$ is finite and even, $|Y\setminus A|$ is finite and odd, and $h\upharpoonright(Y\setminus A)$ is a bijection between these two sets, which is absurd. $\dashv$

Finally, the maps

$$f:Y\to X:y\mapsto\begin{cases} y,&\text{if }y\in X\\ \langle 0,0\rangle,&\text{if }y=p \end{cases}$$

and

$$g:X\to Y:x\mapsto\begin{cases} x,&\text{if }x\in\omega^*\\ p,&\text{if }\pi(x)=0\\ \langle n-1,i\rangle,&\text{if }x=\langle n,i\rangle\text{ and }n>0 \end{cases}$$

are continuous surjections.

By the way, each of $X$ and $Y$ embeds in the other, so these spaces witness the lack of a Schröder-Bernstein-like theorem for compact Hausdorff spaces and embeddings.

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