Does every arbitrary covariant functor $F:\mathcal{C} \to \mathcal{D}$ maps an iso into an iso??

Question

Let $F:\mathcal{C} \to \mathcal{D}$ a covariant functor between two categories I claim that if $f:X \to Y$ is an isomorphism in $\mathcal{C}$ then $F(f)$ is an isomorphism.

As $f:X \to Y$ is an isomorphism in $\mathcal{C}$ then there is another morphism $f':Y \to X$ in $\mathcal{C}$ such $f'f=1_{X}$ and $ff'=1_{Y}$. Then

$$F(f)F(f')=F(ff')=F(1_{Y})=1_{F(Y)},$$

and

$$F(f')F(f)=F(f'f)=F(1_{X})=1_{F(X)}.$$

Proving $F(f):F(X) \to F(Y)$ is an isomorphism in $\mathcal{D}$.

Is this right? Im surprised is just needed just the definition to prove this and every functor preserves isos.

Answer 1

Yes.

If we think about algebraic topology, where functors first arose, we should be able to see that this basic property is very important. That is, isomorphic (homeomorphic) topological spaces would need to have isomorphic homology groups and homotopy groups, as well.

For the attached groups are used to categorize the (topological) spaces.

Answer 2

You're right and the proof is correct. It follows because an isomorphism is just an arrrow with an inverse and functors preserve composition and identities, and hence map the inverse to the inverse. This works in basically any "algebraic" morphism setting that you already know: a ring morphism maps units to units too etc. Categories are sort of "algebraic" in flavour: composition as operation and functors preserve the operation and the identities..