# Tag Info

6

The duality is that between extremal monomorphisms and extremal epimorphisms. A monomorphism is the right generalization of injective function to situations with more structure, such as topological spaces with continuous functions. We say $f:X\to Y$ is a monomorphism if, whenever $g,h:Z\to X$ are such that $f\circ g=f\circ h$, we can infer $g=h$. ...

5

Maybe you've seen a theorem stating that the inverse image of an open subset by a continuous function is an open subset (this is actually a general definition for continuity) ? Here, $A=f^{-1}((-1,1))$.

4

Nope. $A_1 \times B_1 \cup A_2 \times B_2 \neq ( A_1 \cup A_2 ) \times ( B_1 \cup B_2 )$ - a sum of two rectangles usually isn't a rectangle. For longer products/unions it doesn't work either. For this to be correct, you would have to take $k = \max \{ k_{\alpha} : \alpha \in A \}.$ But what if it is infinite?

2

2.3.29. Proposition. The Cartesian product $f=\prod_{s\in S} f_s$, where $f_s \colon X_s \to Y_s$ and $X_s\ne\emptyset$ for $s\in S$, is open if and only if all mappings $f_s$ are open and there exists a finite set $S_0\subset S$ such that $f_s(X_S)=Y_S$ for $s\in S\setminus S_0$. Proof. From 1.4.14 and the equality $f(\prod_{s\in ... 2 Here's another example: $$X=(\mathbb R\times\{0\})\cup(\mathbb Q\times\mathbb Q).$$ Any neighborhood of any point is disconnected, but$\mathbb R\times\{0\}$is a connected component, so the space is not totally disconnected. 2 How about this space? $$\{(x,y)\in\Bbb R^2\mid x\in\{\frac1n+\frac1m\mid n,m\in\Bbb N\}, y\in[0,1] \}$$ 2 I presume you're asking how to equip a given space with a CW-structure. There is no general procedure to do it. One usually tries to express his space into smaller and simpler looking spaces, and then obtain a CW-structure on the whole space from "patching-up" the CW-structure on those smaller spaces. For example, Given a pair of CW-complexes$(X, A)$with ... 2$\bullet \space \mathbf{SO(3) / SO(2) \simeq S^2}:$Consider a fundamental representation of the Lie group$G := SO(3)$. Any element$M$of$G$can be written as a linear map$M : \mathbb{R}^3 \rightarrow \mathbb{R}^3$such that$M^{-1} = M^T$and$\det(M) = 1$. We can easily restrict to$M : S^2 \rightarrow S^2$. For any arbitrary$x \in S^2 \subset ...

1

This is an issue if you don't allow the empty set to be a topological space (and if you do allow it, then there might be issues elsewhere). But generally if the product is empty, then it is empty. Note that this has nothing to do with the particular topology (box, or otherwise). If the entire product is empty, then every subset of it is also empty, because ...

1

Look at an example. The topology of $[0,1]\subset \mathbb{R}$ has as open sets the intersections of open sets of $\mathbb{R}$ and $[0,1]$. So for instance $(1/2,3/4)$, $(3/4,1]$ and $[0,1)$ are open sets of $[0,1]$ in the topology induced by the topology if the real line.

1

It's also often called the subspace topology: http://en.wikipedia.org/wiki/Subspace_topology It provides a subset of a topological space with a topology of its own, and it works in the way you might expect. The subspace topology on $S\subseteq X$ is the one in which a subset of $S$ is open iff it is the intersection of an open subset of $X$ with $S$. In ...

1

Induced basically means it is generated by another through some mean. In this case the subspace $S$ has a topology being generated by the topology of $X$ through the process of intersection.

1

I'm not sure how familiar you are with the box/product topologies, so I'll keep it brief. I'll let Munkres fill in any gaps! EDIT response: If the collection of spaces is finite, then the box and product topologies will agree, so the 'induced' topology of the subspace will still have open $U \subset \prod_i X_i$ such that $X_n \cap U$ is open for some ...

1

For all $x \in A$ you have $-1 < \lim_{t \to x}f(t) <1$, so there exists some $\varepsilon > 0$ such that $(x-\varepsilon, x + \varepsilon) \subseteq A$. This shows that $A$ is open.

1

I assume that by "domain" we mean an open, connected subset of $\mathbb{R}^n$. Note, in particular, that if $D$ is a domain and $K\subseteq D$, then $K\cap \partial D=\varnothing$. Let's solve this using the contrapositives. First suppose that $f$ is not proper. Then there exists a compact set $K\subseteq D_2$ such that $f^{-1}(K)$ is not compact. But since ...

1

Yep! The product topology makes projections continuous. If $C$ is compact then every one of its projections $\pi_x(C)$ is compact, and $C\subset\prod\pi_x(C)$.

Only top voted, non community-wiki answers of a minimum length are eligible