I know it's bad habit studying wikipedia as a proper source. But I am particularly troubled by the statement that: "In concrete categories, a function that has a left inverse is injective. Thus in concrete categories, monomorphisms are often, but not always, injective. The condition of being an injection is stronger than that of being a monomorphism, but weaker than that of being a split monomorphism." when I couldn't tell the differences between "left invertible" and "split monomorphisms" Or is injective a legal concept in category theory?
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A morphism being 'injective' is a valid concept in concrete categories, because by definition those categories have the additional structure of 'underlying sets', ie they come equipped with a faithful functor into the category of sets. The bolded statement says that there are three separate notions:
- Being a monomorphism when considered as a map in your category.
- Being an injection, ie the image of your map under the aforementioned faithful functor is injective as a map of sets.
- Being a split monomorphism (which is the same as being left-invertible)
Here (3) implies (2) implies (1)
answered Oct 29, 2017 at 20:05
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$\begingroup$ I have a little question on monomorphism but I don't think its worth to be asked as a normal question, I was thinking about a proof for $f(x)=x$ if $f$ is a monomorphism. Can you please help by giving me a proof for that? $\endgroup$ Nov 27, 2018 at 7:11
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$\begingroup$ @FareedAF Identity maps are always trivially monomorphisms (since $\text{Id}\circ f = \text{Id}\circ g$ is equivalent to $f = g$) if that's the implication you're trying to prove, but the converse is false. $\endgroup$ Nov 27, 2018 at 15:47
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1$\begingroup$ @FareedAF The existence of non-identity automorphisms (eg complex conjugation as an $\Bbb R$-automorphism) provide counterexamples. $\endgroup$ Nov 27, 2018 at 19:35
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1$\begingroup$ @FareedAF I see. Your initial comment implicity had $f(x) = x$ for all $x$ (which is false for non-real elements of $\Bbb C$ in this example, eg $f(i)=-i$), but when you specify $f(x) = x$ only for all $x$ in the base field $K$, then that's defintionally fine for maps of fields that fix the base field $K$ (if that's your definition of $K$-monomorphism. if you have a different definition of "$K$-monomorphism" then there's a possibility of being tripped up by the distinction between isomorphism and identity) $\endgroup$ Nov 27, 2018 at 19:57
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1$\begingroup$ @FareedAF I would expect the definition of $K$-monomorphism to include a reference to $K$ somewhere. By what you've said the only way I can implicitly do this is to interpret the word "homomorphism" to mean "homomorphism of $K$-algebras" (as opposed to just 'homomorphism of fields'), which would again definitionally imply that $f(k)=k$ for $k \in K$. $\endgroup$ Nov 28, 2018 at 6:06