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\begin{align} f: X \longrightarrow Y \\ g: Y \longrightarrow Z \\ g \circ f: X \longrightarrow Z \tag{is bijective} \end{align} The bijective conditions only applies to $ g \circ f$. I already managed to show that in this case $f$ must be injective too, I did it the following way (maybe you can check if I did it correct):

Suppose: $f$ is not injective, then $x, x' \in X$ such that $f(x)=f(x') , \ x\neq x'$. For the composition this means that $(g \circ f) (x) = (g \circ f)(x') \implies x=x'$ because the composition is by definition a bijection and this statement would lead to a contradiction. Thus $f$ is a injection.

I am content so far with this proof, however if I want to analyze the properties of $f$ whether it's also surjective or not I get stuck.

Question: is $f$ surjective?

I assume that $f$ is also surjective and therefore a bijection. This means that $\forall y \in Y \exists x \in X : f(x)=y$, I also know that the composition is surjective, because it's bijective, this means that $\forall z \in Z \exists x \in X: g(f(x))=z$

So I can say that $g(f(x))=z=g(y)$ Is this a contradiction I should notice here? Because I have read (also from other posts on here) that $f$ must not necessarily be surjective, but I couldn't come up with a complete proof on my own so far.

Edits: Improved formatting as suggested by @Thomas Andrews, corrected a mathematical false statement as suggested by @Michael Albanese

Spaced
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  • Can you actually state the problem? You have $f,g$ and you know $g\circ f$ is bijective. What are you trying to answer? (We should not have to refer back to the title to get that information.) – Thomas Andrews Sep 27 '13 at 16:21
  • $f\colon\mathbb{Z}\rightarrow\mathbb{Z}$ given by $f(n)=2n$ and $g\colon\mathbb{Z}\rightarrow\mathbb{Z}$ given by $g(2n)=n$, $g(2n+1)=0$ are functions satisfying $g\circ f$ is bijective. $f$ is not surjective, $g$ is not injective. – Dan Rust Sep 27 '13 at 16:22
  • Your statement of what it means for $f$ not to be injective is incorrect. It reads "if $f(x) = f(x')$ then $x \neq x'$"; what about the case $x = x'$? Instead say that there are $x, x' \in X$, $x \neq x'$ such that $f(x) = f(x')$. – Michael Albanese Sep 27 '13 at 16:22

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Assuming $f$ is a surjective map will not lead to a contradiction because it is possible for $f$ to be both injective and surjective if $g \circ f$ is a bijection. An example of this is where $g$ is the identity map. If $f$ is surjective, then $f$ is a bijection so $f^{-1}$ exists and is itself a bijection. As compositions of bijections are bijections, $g = (g\circ f)\circ f^{-1}$ is a bijection.

The point is that $f$ doesn't have to be surjective (so $g$ doesn't have to be a bijection); there are examples where $f$ is not surjective but $g\circ f$ is a bijection. One such example is $f: \{0\} \to \{0, 1\}$, $f(0) = 0$, is not surjective but for $g : \{0, 1\} \to \{0\}$, $g(0) = g(1) = 0$, the composition $g \circ f : \{0\} \to \{0\}$, $(g\circ f)(0) = 0$ is a bijection.

A useful exercise in trying to understand the situation is to determine what restrictions are imposed on $f$ and $g$ by requiring $g\circ f$ to be a bijection. You've already determined one of them, namely that $f$ has to be injective, but this alone does not guarantee $g\circ f$ will be a bijection.

Michael Albanese
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