29 votes
Accepted

Example of a functor that doesn't reflect isomorphism

A very generic counterexample: Take for $\mathcal{D}$ the category with one object $O$ and one morphism $f$. There is a functor $F$ from any category to $\mathcal{D}$ that sends every object to $O$ ...
Ward Beullens's user avatar
20 votes
Accepted

Can the map sending a presentation to its group be considered as a functor?

Yes, yes, yes, to all three questions. And it can be done very generally and very nicely for universal algebra $\ $ [or even for first order structures]. Let's fix an algebraic signature consisting of ...
Berci's user avatar
  • 90.8k
19 votes
Accepted

What is the difference betwen equivalence and isomorphism of functors in categories.

Well, the actual difference between the two statements is that for an equivalence of categories, we only require that that the composites $F \circ G$ and $G \circ F$ are naturally isomorphic to the ...
Chessanator's user avatar
  • 5,458
15 votes

Why is a natural transformation not a functor of functors?

A functor from $\mathcal A$ to $\mathcal B$ must take every morphism in $\mathcal A$ to a corresponding morphism in $\mathcal B$. A natural transformation connects one particular functor $\mathcal A\...
hmakholm left over Monica's user avatar
15 votes
Accepted

Does the vanishing of the first derived functor really imply a functor is exact?

Yes, this is correct. The point is that the vanishing of the first derived functor on all objects is a very strong condition, and that the first derived functor on one object will correspond to ...
Eric Wofsey's user avatar
13 votes

Example of a functor that doesn't reflect isomorphism

Just take the forgetful functor $F$ from $\mathit{Top}$ to $\mathit{Set}$. Then, take a bijection $f$ between two topological spaces which is not a homeomorphism. Of course, $Ff$ will be an ...
José Carlos Santos's user avatar
13 votes

What constructions of "elementary" mathematics are actually functors?

One of my favourite examples of these is group actions. A (monoid or) group $G$ can be considered as a category with a single object $\star$, whose morphisms $\star \to \star$ are the elements of $G$,...
Clive Newstead's user avatar
13 votes

Does every functor from Set to Set preserve products?

There are a lot of counterexamples, actually. For example, if $S$ is has more than two elements, then the functor that map every set to $S$ and every function to the identity of $S$ does not preserve ...
Arnaud D.'s user avatar
  • 20.9k
13 votes

Counterexample: functor does not preserve monomorphisms

The functor $F^{op}$ does not preserve monomorphisms.
Oskar's user avatar
  • 4,363
13 votes
Accepted

Are fully faithful functors stable under pullback?

There is a bijective-on-objects/fully faithful orthogonal factorisation system on Cat. Hence, as a right class of a factorisation system, fully faithful functors are closed under pullback in Cat. For ...
varkor's user avatar
  • 5,683
12 votes
Accepted

Quasi-inverse of an equivalence of categories is unique up to unique isomorphism

It's simply not true: the isomorphism is not unique. Indeed, you could have $G=G'$, in which case the claim is that any functor which is an equivalence has no automorphisms besides the identity. ...
Eric Wofsey's user avatar
12 votes
Accepted

Can functoriality be explained in the opposite way?

In general, if all you know is that $F(f)\circ F(f')=F(f'')$, the composition $f\circ f'$ may not even be defined, and even if it exists it need not coincide with $f''$. For example, if $X,Y,Z$ are ...
Arnaud D.'s user avatar
  • 20.9k
12 votes

Can the map sending a presentation to its group be considered as a functor?

Giving a presentation $\langle S \mid R \rangle$ of a group amounts to describing it as the cokernel of a map $F(R) \to F(S)$ between free groups. There is a category $C$ whose objects are such maps ...
Qiaochu Yuan's user avatar
12 votes
Accepted

Any two natural transformations between identity functors commute

Consider the diagram $\require{AMScd}$ \begin{CD} C @>{\alpha_C}>> C \\ @V{\beta_C}VV @VV{\beta_C}V \\ C @>{\alpha_C}>> C. \end{CD} Regard $\beta_C$ as just any morphism. Apply the ...
Арсений Кряжев's user avatar
11 votes

Example of a functor that doesn't reflect isomorphism

Homology and homotopy functors are very natural examples of this.
Pedro's user avatar
  • 122k
11 votes

Counterexample: functor does not preserve monomorphisms

The reason why it is more "difficult" to find a non mono-preserving functor in nature is that forgetful functors $\mathbf C \to \mathsf{Set}$ preserve (even create) finite limits when $\mathbf C$ is a ...
Pece's user avatar
  • 11.6k
10 votes
Accepted

Functor of points of the affine line with double origin

Let $X$ denote the affine line with doubled origin, which is covered by two affine lines $A$ and $B$ which are identified everywhere except the origin. Then the functor of points of $X$ can be ...
Eric Wofsey's user avatar
10 votes
Accepted

Is power set functor determined by its image on objects?

There exists at least one other endofunctor of $\mathbf{Set}$ that sends every set to its powerset. This endofunctor sends a function $f:X\to Y$ to $$\widehat{f} :P(X)\to P(Y):U\mapsto \widehat{f}(U)=\...
Arnaud D.'s user avatar
  • 20.9k
10 votes
Accepted

Prove that $\mathbf{FinSets}^{\mathbf{N}}$ has no subobject classifier.

There's a standard trick using the Yoneda lemma for computing what universal objects in (restricted) functor categories must be if they exist. In the case of the subobject classifier, this is ...
Alex Kruckman's user avatar
10 votes

A category of functors?

A natural transformation between functors $F,G:I \rightarrow C$ is nothing but a family of maps $(Fi \rightarrow Gi)_{i\in \operatorname{Ob}I}$ satisfying some compatibility conditions. Hence the ...
Jonas Linssen's user avatar
9 votes
Accepted

Algebraic topology for non-nice spaces

Actually, there is an active theory of algebraic topology for "pathological" spaces that has come a long way in the past two decades: Wild (algebraic/geometric) Topology. In fact, this has become a ...
Jeremy Brazas's user avatar
9 votes
Accepted

Product in the category of functors.

Yes, the product of $F_i\in\text{Obj}(\text{Fun}(A,\mathbf{Set}))$, $i\in\Lambda$, is the functor $\prod_{i\in\Lambda}F_i$, such that $(\prod_{i\in\Lambda}F_i)(X)=\prod_{i\in\Lambda}(F_i(X))$ for ...
Oskar's user avatar
  • 4,363
9 votes
Accepted

Does every functor from Set to Set preserve products?

If you consider the powerset functor $\mathcal{P}$, which takes any $A$ to its powerset $\mathcal{P}A$, and any function $f : A \to B$ to the function $\mathcal{P}f : \mathcal{P}A \to \mathcal{P}B$, ...
Thibaut Benjamin's user avatar
9 votes
Accepted

Bifunctoriality stronger than functoriality in each variable?

Yes, bifunctoriality is a stronger condition than functoriality in each variable, and yes, the commutative diagram you write down need not commute in the second case. I don't actually know an example ...
Qiaochu Yuan's user avatar
8 votes
Accepted

End of Hom-profunctor in Grp

You're either approaching, or already made without saying so explicitly, the realization that the end of the hom bifunctor is the set of natural endomorphisms of the identity functor. This follows ...
Kevin Carlson's user avatar
8 votes

Example of a functor preserving only finite coproducts

The following example seems to work. Consider the composition $$F : \mathsf{Set} \xrightarrow{D} \mathsf{Top} \xrightarrow{\beta} \mathsf{CompHaus} \xrightarrow{U} \mathsf{Set},$$ where $D$ is the ...
8 votes
Accepted

What is the difference between the identity functor and the identity morphism?

If $X$ is an object in a category $\mathcal C$, the identity morphism $id_X:X\to X$ is a morphism in the category $\mathcal C$. On the other hand, the identity functor is not a morphism in $\mathcal ...
Alex Mathers's user avatar
  • 18.5k
8 votes
Accepted

Forgetful functor from complete metric space to metric space

This forgetful functor doesn't really "forget" any structure the way most forgetful functors do: it just forgets the fact that your metric spaces happen to be complete. This doesn't change the ...
Eric Wofsey's user avatar
8 votes
Accepted

Functor to the empty (objectless) category?

There is indeed a category $E$ with no objects (and therefore no morphisms). However, if a category $A$ has at least one object then there does not exist any functor $A\to E$. Let's think about sets ...
Tashi Walde's user avatar
  • 1,526

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