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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.


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.

  • $\begingroup$ Let me ask you couple questions: 1) Why we always have $T(\overline{A})\subset \overline{T(A)}$? 2) Why $Q^n=\overline{\text{int}Q^n}$? $\endgroup$ – ZFR May 4 '16 at 16:56
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    $\begingroup$ 1) 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 $F\subseteq Y$ $\endgroup$ – Daniel Fischer May 4 '16 at 17:36
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    $\begingroup$ (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 $T(x) \in \overline{T(A)}$. $\endgroup$ – Daniel Fischer May 4 '16 at 17:37
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    $\begingroup$ 2) 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 $B_{\varepsilon}(\mathbf{u}) \subset Q^n$). $\endgroup$ – Daniel Fischer May 4 '16 at 17:37
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    $\begingroup$ 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. $\endgroup$ – Daniel Fischer May 4 '16 at 17:37

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