If $f \circ g$ is injective and $g$ is surjective - is $f$ injective?
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If $g \circ f$ is injective, so is $g$
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Show that $g$ is injective
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What can you say about $f$ and $g$ in the case that $fg$ is 1) Injective, 2) Surjective - Cohn - Classic Algebra Page 15
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Finite order of an element $f(g)$ of a homomorphism $f$
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proof verification: If $f:G \rightarrow H$ is group homomorphism, and $H$ is abelian, then $G$ is abelian
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If $g \circ f$ is surjective, show that $f$ does not have to be surjective?
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Proving that $f(x) = g(x)$ for some $x \in [a,b]$ if $f,g$ continuous, $f(a)  g(b)$
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How to prove that the composition of two surjective functions is surjective
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Constructing a Bijective function via an injective and surjective one.
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A function $f\colon E\to F$ between finite sets is injective iff it is surjective iff it is bijective
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Characterising functions $f$ that can be written as $f = g \circ g$?
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$g\circ f$ is Surjective and $g$ is Injective then Prove that $f$ is surjective
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$f(x)=x/1$ is not injective?
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Groups $G,H$ with surjective homomorphisms that are quotient of each other
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If $\operatorname{ran} F \subseteq \operatorname{dom} G$, then $\operatorname{dom}(G \circ F) = \operatorname{dom} F$.
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If $\forall x \in R, x^2-x \in Z(G)$, than $R$ is commutative
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Let $f:K\to K$ with $\|f(x)-f(y)\|\geq ||x-y||$ for all $x,y$. Show that equality holds and that $f$ is surjective.