I'm currentlly doing a course in abstract algebra and I often have to prove a map is surjective or injective. It's always done the same way, we take $f(a)=f(b)$ and deduce $a=b$, or we show that for every $y$ in the range there is an element x in the domain such that $f(x)=y$. I was wondering if there are alternative ways we can use to prove a map is injective/surjective?
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3You can prove that it is bijective by demonstrating an inverse. – Patrick Stevens Nov 17 '18 at 19:04
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You can try to show that $f$ is a composition of some injective/surjective functions. Anyway I vote to close this as being too broad. – Yanko Nov 17 '18 at 19:05
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If the map is an arbitrary function, there is no better way.
However, if the map is a homomorphism, you can prove it is injective by showing its kernel is trivial.
ATOMP
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1I don't agree that this is the only way (see my answer). However the comment about checking kernel is very useful in practice. Similarly surjectivity may sometimes be checked by computing the cokernel. Be warned though that cokernels don't always exist (it depends on the type of algebraic structures you are working with). – Blazej Nov 17 '18 at 19:11
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Suppose that $X,Y,Z$ are sets and $f: \ X \to Y$ and $g : \ Y \to Z$ are functions. We have the following implications: if $g \circ f$ is a surjection, then $g$ is surjection. If $g \circ f$ is an injection, then $f$ is an injection. I encourage you to prove this lemma yourself if you haven't seen it before.
Blazej
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This was very helpful, thank you. I vaguely remembered it from class but I knew there was something similar to this. – Lowkey Nov 17 '18 at 19:54