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5

Define $F(x,t)=(1-t)x$. Note that it is not true that a space is necessarily contractible if it has trivial fundamental group. For example, any sphere of dimension greater than 1 has trivial fundamental group and is not contractible.


4

In this case $F$ is injective : if $F(x) = F(y)$ then $0 = d(F(x),F(y) \geq L d(x,y)$ implies that $d(x,y) = 0$, that is, that $x=y$. As $F$ is surjective, it is bijective, and its inverse $F^{-1}$ is Lipschitz with coefficient $1/L < 1$, so as $X$ is complete, this inverse $F^{-1}$ has a fixed point $x_0$, which is also a fixed point for $F$. Finally, ...


1

$T^3$ is a Lie group,so your space $T^3\times T^3-\Delta$ is a trivial bundle over $T^3$ with fiber $T^3-\{\ast\}$ (google"configuration space" for more details).So $$\pi_1(T^3\times T^3-\Delta)\cong \pi_1(T^3\times (T^3-\{\ast\}))\cong Z^3\times Z^3\cong Z^6$$


1

This property has been called hypocompactness; it is strictly weaker than compactness and strictly stronger than paracompactness. The Sorgenfrey line is hypocompact but not compact: every open cover has a disjoint clopen refinement. Any hedgehog space $X$ of uncountable spininess is an example of a paracompact space that is not hypocompact. $X$ is ...


1

Here are some hints: (Whenever I speak of a map betweens spaces, I'll mean a continuous function.) Note that the compact subsets of $X$ are exactly the images $t(C)$ of maps $t:C→X$ where $C$ is a compact space (call such a map a test map). Then $X_c$ has the final topology for all test maps to $X$, which are the same as the test maps to $X_c$, so $X_c$ is ...


1

The intuitive idea that topology captures is the idea of separation (in some sense) of points and subsets of a set. The intuition is most obvious in the most basic (least interesting and most extreme) topologies; the discrete topology (the collection of all subsets of a given set) where every two points and every two subsets can be separated and the ...


1

Let $\tau$ be any topology on $X$ with the stated property; what you’re being asked to prove is that $\tau\supseteq\mathcal{T}$, which amounts to proving that the sets $\{x\in X:x<a\}$ and $\{x\in X:a<x\}$ belong to $\tau$ for each $a\in X$. For convenience let $(\leftarrow,a)=\{x\in X:x<a\}$ and $(a,\to)=\{x\in X:a<x\}$. HINT: Fix $a\in X$. For ...


1

HINT: Let $\tau$ be the order topology on $Y$, and let $\tau_Y$ be the subspace topology on $Y$ that it inherits from $X$. You’re asked to show that $\tau\subseteq\tau_Y$, and that there are examples in which $\tau\subsetneqq\tau_Y$. Showing that $\tau\subseteq\tau_Y$ is straightforward: you need only show that for each $y\in Y$, $\{z\in Y:z<y\}$ and ...


1

What i understood is that your are trying to find a homotopy of $id_{S^1}$ to $a:S^1\to S^1$ such that $a(x)=-x$. This is the antipodal map. Such a homotopy is $F:S^1\times I\to S^1$ with $F(x,t)=e^{i\pi t}x$. It's is clearly continuous, $F(x,0)=x$ and $F(x,1)=-x=a(x)$. Also the projective plane $\mathbb RP^2$ is the quotient space $S^1/_{x\equiv a(x)}$


1

$Y_1$ is indeed open, in both $X$ and $\mathbb{R}$, but not for the reason you say. It's open, because every point is an interior point, or you could look at the complement, which in $\mathbb{R}$ is $(-\infty,1] \cup [2, \infty)$, which is indeed closed (it contains all its limit points). But openness is more direct, as $Y_1$ is an open interval which is ...



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