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6

Regarding the duality between limits and colimits, what you're overlooking is that you can also take the opposite of the index category as well. It may be helpful to observe that cones and cocones are morphisms in the functor category $C^J$. If $J$ is the index category, then there is the 'diagonal' embedding $ \Delta : C \to C^J $ that sends every object ...


6

Yes! It is the Sierpinski Space. You will find most answers in the wikipedia link. It is a connected two point set and is really useful for plenty counterexamples and/or constructions. From a categorical viewpoint the Sierpinski Space represents the functor $X \mapsto \tau (X)$, $f\mapsto f^{-1}$.


5

For every object $ A $, the arrow $ A\longrightarrow 0$ is a split epimorphism. Therefore (c) says that this arrow is an isomorphism, which implies (d).


4

The usual closure operator on subsets of topological spaces doesn't satisfy the last condition: in general, $f^{-1}(\overline{S})$ can be strictly larger than $\overline{f^{-1}(S)}$. For example, $f$ might have the property that its image doesn't intersect $S$ but does intersect the closure of $S$.


3

You want to look at the category of diagrams $$ 1 \to A \to A $$ where the two copies of $A$ should be interpreted as being the same object, rather than two separate objects of the index category that happen to map to the same object.


3

Suppose $f:X\to Y$ is such that $f(X)$ is not dense. That is, let $B(x,r)$ be an open ball in $Y$ which $f(X)$ does not intersect. Let $g_1:Y\to \mathbb{R}$ be $g_1(y)=0$, and $g_2:Y\to \mathbb{R}$ be $g_2(y)=\max(r-d(y,x),0)$. The maximum of continuous functions is continuous, so the $g_i$ are continuous functions which agree on $f(X)$ but don't agree on ...


3

Hint. Using $X \simeq X \times 1$ and $1 \simeq 0$, you can prove $$ \mathcal C(X,Y) \simeq \dots \simeq \{\ast\}. $$ From there, you can deduce that any object is both initial and final.


2

An NNO is an initial object in the category of algebras for the endofunctor $X \mapsto 1+X$.


2

Here is the answer that I wrote a few months ago, which should pertain to this question as well: Both categories are denoted by $\mathrm{Rel}$. In most cases, however, the author either explicitly tells which $\mathrm{Rel}$ he/she is using, or reproduces the definition of $\mathrm{Rel}$. (Note: this applies to any definition/term in math which can be ...


2

Let $A_1,\ldots,A_n$ be objects, and let $P$ be the product, with projections $\pi_1,\ldots,\pi_n$. For every $j=1,\ldots,n$ define $i_j:A_j\to P$ to be the unique morphism that satisfies $\pi_j\circ i_j=id_{A_j}$ and $\pi_k\circ i_j=0$ for every $k\neq j$. ($i_j$ is well defined for every $j$ by the universal property of the product). We show that $P$, ...


1

It's useful to know that in order to show that a function $g:Y\to X$ is continuous, it suffices to show that for each $S\in\cal S$, where $\cal S$ is a subbase, the preimage $g^{-1}(S)$ is open. Remember that by taking all finite intersections of element in $\cal S$ we obtain a base $\cal B$ for the topology on $X$, and each open set is then an arbitrary ...



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