# Tag Info

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In general, in a category $\mathsf{C}$ that has products, an object $c$ such that $- \times c : \mathsf{C} \to \mathsf{C}$ has a right adjoint is called exponentiable. There are various characterizations of exponentiable objects in $\mathsf{Top}$, and you can find some of them in this $n$Lab article: Theorem (Exponentiability, I). An object $X$ of ...

7

An endomorphism is a morphism from an object to itself, full stop. I don't know why you're so distrustful of people who aren't category theorists (I'm a graduate student who studies category theory myself; is that good enough or do I need a PhD first?), but, for example, this is the definition you will find on Wikipedia, the nLab, a different page of ...

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Surjectivity has nothing to do with being endomorphism. Think for example a constant map from a set of more elements to itself: that's a not surjective endomorphism in the category of sets and functions.

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If $A$ is any ring, then $A$ and $M_n(A)$ have equivalent categories of modules, and usually $A$ and $M_n(A)$ are not isomorphic. This is the simplest example of a Morita equivalence.

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$\mathbb{Q}$ is a Hausdorff space. Every Hausdorff space is sober, so its soberification is just itself.

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An explicit counterexample is $\text{Ab}^{op}$ (an excellent counterexample in general), which by Pontryagin duality is equivalent to the category of compact (Hausdorff) abelian groups. The product is what you think it is but the coproduct is more complicated: it is not the direct sum but the Bohr compactification of the direct sum, and because this ...

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An endomorphism is a morphism which has the same domain and co-domain. These "morphisms" do not have to be functions at all. The morphisms don't have to have any meaning behind them. All that matters is that for every morphism we assign to it a domain and a co-domain. A morphism in which the domain and co-domain are equal is called an endomorphism.

5

You're working too hard. I'll assume you're convinced that $\mathbb{Z}$ is the free group on one generator, which means you're convinced that $$\text{Hom}(\mathbb{Z}, G) \cong G$$ (where the RHS should really be the underlying set of $G$ but I'm omitting that I'm applying the underlying set functor). One way to state the universal property of the coproduct ...

4

Spanier's book is relatively old (so I know it does not quite answer your question), but excellent. It uses category theory from the get-go. Riehl's "Categorical homotopy theory" is very well-written, though it may be a bit too advanced if you hadn't seen a bit of algebraic topology already. Riehl's book is focused on the categorical aspect via Quillen model ...

3

You only need to understand what the first dual map $\phi^\vee$ is, as $(\phi^\vee)^\vee=\phi^{\vee\vee}$. The map $\phi^\vee$ is defined by $\phi^\vee(f)=f\circ\phi$. This is just a special case of the contravariant hom functor. To check that the diagram is commutative, notice that for $v\in V$ and $f\in W^\vee$ we have $$(\phi^{\vee\vee}\circ ... 3 Yes, absolutely. The approach I know about is synthetic differential geometry, which begins with thinking of manifolds not via their underlying locally Euclidean topology (this is analytic differential geometry!) but via their smooth functions, in particular those to and from the line R (R behaves differently enough from \mathbb{R} that it's a good ... 2 If you want to write a map S : X \times Y \to Y \times X, you could use the UPP for Y and X: given two morphsims f : A \to Y and g : A \to X, there exists an unique morphism h = \langle f , g \rangle : A \to Y \times X such that \pi_Y \circ h = f and \pi_X \circ h = g. To find S just take A = X \times Y, f = \pi_Y and g = \pi_X, and ... 2 I guess (from the comment discussion mostly) your question is not really about free objects but rather: given an adjunction F \dashv U, how can I explicitly write the bijection \hom(FA, B) \to \hom(A, UB) so that the special case \mathsf{Set} \leftrightarrows \mathsf{Grp} gives me the restriction \varphi \mapsto \varphi\restriction A? Given F ... 2 If X\stackrel f\leftarrow A\stackrel g\to Y is a cotriad, the pushout is$$X\stackrel{\bar g}\to Z\stackrel{\bar f}\leftarrow Y where $Z$ is the quotient space of $X\sqcup Y$ by the relation generated by $f(a)\sim g(a)$, and $\bar f$ is the composite $Y\to X\sqcup Y\to Z$. As often with relations, one only says what generates it, and the whole relation ...

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There is no neighborhood filter in $\mathbb{Q}$ of $x\in\mathbb{R}\setminus\mathbb{Q}$, at least not one satisfying the correct axioms. In $\mathbb{Q}$, we have $(x-1,x+1) = (x-1,x)\cup(x,x+1)$. So the filter of open sets containing $x$ cannot be prime, and should be illegal in whatever source you're using.

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In principle, "$F$ is a left adjoint" is fine because the adjunction is determined uniquely up to unique isomorphism by $F$. This seems to contradict your intuition that "$F$ is a left adjoint" sounds wrong, but it's perfectly consistent with standard usage such as talking about "the" limit of $G$: this sort of usage works fine except in contexts where it's ...

1

The straightforward answer to your question is: sure, but they're not very interesting. If we view a (2-valued) model $M\models ZF$ (I use $ZF$ for simplicity) as a category whose objects are sets in $M$ and whose morphisms are functions in $M$, as you do, then - whenever we have an end-extension $N\supseteq M$ - the inclusion map is a functor from $M$ to ...

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No. Consider the functor $\Delta : \mathbf{Set} \to \mathbf{sSet}$ that sends each set $X$ to the discrete simplicial set on $X$. This is certainly fully faithful, and it has a left adjoint, namely the functor $\pi_0 : \mathbf{sSet} \to \mathbf{Set}$ that sends a simplicial set to its set of connected components. It is not hard to check that $\pi_0 : ... 1 First note that a monomorphism$f:X\to Y$must be injective: If$f(x)=f(y)$, then the compositions of$f$with the maps$(*\to X)$sending the single point to$x$and$y$, respectively, are equal, thus these maps are equal, hence$x=y$. So let$f:X\to Y$be a monomorphism in$\mathbf{CHaus}\$. Since a map from a compact space to a Hausdorff is perfect ...

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The closest thing I've found is Strom's Modern Classical Homotopy Theory, although I haven't read much of it. Chapter 1 is called Categories and Functors, so that's a good start. This is the only introductory algebraic topology textbook I know of that explicitly uses the language of homotopy limits and colimits.

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Rotman's An Introduction To Algebraic Topology is a great book that treats the subject from a categorical point of view. Even just browsing the table of contents makes this clear: Chapter 0 begins with a brief review of categories and functors. Natural transformations appear in Chapter 9, followed by group and cogroup objects in Chapter 11. The aspect I ...

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I like the picture of a cocone from http://arxiv.org/abs/math/0306223: They describe it through a story about people sending emails to each other. And then: A cocone which is universal is a colimit. ―http://ncatlab.org/nlab/show/cocone Added: Colimits glue. Limits cut. ―Adam Hughes

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