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How does he use the inverse function theorem (IFT) in proving that $T(Q^n)=\overline{U}$ for some open set $U$ in $\mathbb{R}^n$?

From IFT we conclude that $T$ is an open mapping, hence $T(\text{int}Q^n)$ is also an open set. I guess that he then uses the following property: $T(\overline{A})=\overline {T(A)}$. Am I right or not?

Also why did he write this since he does not use it nowhere?! This seems weird to me.

giobrach
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RFZ
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1 Answers1

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From IFT we conclude that $T$ is an open mapping, hence $T(\text{int}Q^n)$ is also an open set.

That's right, and is an important part of the proof.

I guess that he then uses the following property: $T(\overline{A})=\overline {T(A)}$.

Here we have to be careful. Generally one does not have that equality. If $T$ is continuous, then we always have $T(\overline{A}) \subseteq \overline{T(A)}$, but the inclusion may be strict, one only has equality if $T(\overline{A})$ is closed. However, the simplex $Q^n$ is compact, hence for continuous $T$ the image $T(Q^n)$ is also compact, and since we're looking at Hausdorff spaces, $T(Q^n)$ is thus closed. Therefore, we have

$$T(\operatorname{int} Q^n) \subseteq T(Q^n) = T(\overline{\operatorname{int} Q^n}) \subseteq \overline{T(\operatorname{int} Q^n)},$$

and since $T(Q^n)$ is closed, the last inclusion is in fact an equality.

And now, since $T$ is assumed injective, we can deduce that $T$ maps the topological boundary of $Q^n$ to the topological boundary of $E$, as well as it maps the boundary chain of $Q^n$ to the boundary chain of the differentiable simplex $T$. So in nice situations, the topological notion of boundary and the notion of boundary for differentiable chains are compatible. (The two notions are quite different if e.g. the dimension of the differentiable simplex is smaller than the dimension of the space, then the image of the differentiable simplex has empty interior and coincides with its topological boundary, while the image of the boundary chain is, for non-degenerate simplices, a proper - lower-dimensional - subset of the image of the simplex.)

As to why he made that remark, I don't know. If it's intended to illustrate the compatibility of the topological notion of boundary with the chain-notion (for nice chains), I think some more elaboration would be appropriate.

Daniel Fischer
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  • Let me ask you couple questions: 1) Why we always have $T(\overline{A})\subset \overline{T(A)}$? 2) Why $Q^n=\overline{\text{int}Q^n}$? – RFZ May 04 '16 at 16:56
  • Dear Daniel! I guess that I can answer to the first question abd I'll write proof soon. But the second question is actual! – RFZ May 04 '16 at 17:16
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  • For continuous $T$. That's in fact a characterisation of continuity, a map $f \colon X \to Y$ between topological spaces is continuous if and only if for all $A\subseteq X$ we have $f(\overline{A}) \subseteq \overline{f(A)}$. Depending on which definition of continuity you're used to, different proofs are most appropriate. If your definition is that for every open $U \subseteq Y$ the preimage $f^{-1}(U)$ be open, the relation $f^{-1}(Y\setminus B) = X \setminus f^{-1}(B)$ for all $B\subseteq Y$ shows that $f$ is continuous if and only if $f^{-1}(F)$ is closed for all closed $F\subseteq Y$
    • – Daniel Fischer May 04 '16 at 17:36
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    (this is also sometimes used as the definition). Then from $A \subseteq T^{-1}(T(A)) \subseteq T^{-1}(\overline{T(A)})$ and the closedness of the last set we obtain $\overline{A} \subseteq T^{-1}(\overline{T(A)})$, and that is just $T(\overline{A}) \subseteq \overline{T(A)}$. If your definition is sequential continuity (which is equivalent in metric spaces), then note that for $x \in \overline{A}$ there is a sequence $(a_n)$ in $A$ with $a_n \to x$, and by continuity $T(a_n) \to T(x)$, so $T(x)$ is the limit of a sequence in $T(A)$, and that implies $T(x) \in \overline{T(A)}$. – Daniel Fischer May 04 '16 at 17:37
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  • You can see that for $n \in {1,2,3}$ if you sketch the simplex in these dimensions. Of course seeing is not a proof, but it helps. $\operatorname{int} Q^n = { \mathbf{u} \in \mathbb{R}^n : 0 < u_i \text{ for } 1 \leqslant i \leqslant n\text{ and } u_1 + \dotsc + u_n < 1}$. You can verify that this set consists only of interior points of $Q^n$ (for such a $\mathbf{u}$, choose $\varepsilon < u_i/n$ and $\varepsilon < (1 - u_1 - \dotsc u_n)/n$, then the triangle inequality shows $B_{\varepsilon}(\mathbf{u}) \subset Q^n$).
    • – Daniel Fischer May 04 '16 at 17:37
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    Conversely, you can easily check that the points of $Q^n$ where one coordinate is $0$ or the sum equals $1$ doesn't belong to the interior (if one coordinate is $0$, making that coordinate any smaller takes you out of $Q^n$, if the sum equals $1$ making any coordinate any larger takes you out). With that characterisation, you can see that for every $\mathbf{u} \in Q^n$ there are interior points arbitrarily close to $\mathbf{u}$. Of course we only need to consider points that aren't interior points. – Daniel Fischer May 04 '16 at 17:37
  • For all $i$ with $u_i = 0$, add a small quantity to that coordinate, and if that gets you a sum $\geqslant 1$, subtract a bit from the coordinates where $u_i > 0$ to get the sum below $1$ again. – Daniel Fischer May 04 '16 at 17:38
  • Note that Rudin assumes that $T$ is injective. Why he requires this condition? Is it so important? He does not comment this. – RFZ May 05 '16 at 07:53
  • Rudin says «it is customary and convenient to talk about "oriented boundaries" of certain sets as well. We shall now describe that briefly.» When $T$ is injective, you have a simple relation between $\partial^{\text{top}} T(Q^n)$ and $T(\partial^{\text{top}} Q^n)$ - namely, the two are equal. When $T$ is not injective, then typically $T$ maps part of the boundary of $Q^n$ to the interior of $T(Q^n)$, and so you don't have a good correspondence between the boundary of the set $T(Q^n)$ and the boundary chain. – Daniel Fischer May 05 '16 at 09:28
  • Dear Daniel! I am so sorry that I am worry you with my permanent questions but I would very-very grateful to you if you'll help me with my last question. http://math.stackexchange.com/questions/1772496/unit-square-as-union-of-two-simplexes. I am asking them because Rudin misses IMHO very useful moments and I can't answer to them by myself :( – RFZ May 05 '16 at 17:44
  • Hi, dear Daniel! I am reading the proof of Stokes theorem from Rudin's book and one moment seems to me very confusing. Can you help me with this moment please? http://math.stackexchange.com/questions/1775769/stokes-theorem-from-pma-rudin-confusing-moment-with-simplex – RFZ May 07 '16 at 18:21