Injection and Surjection

• Let $x_1,x_2 \in X$ such that $f(x_1) = f(x_2).$ Then $x_1 = \text{id}_X(x_1) = (g\circ f)(x_1) = g(f(x_1)) = g(f(x_2)) = (g\circ f)(x_2) = \text{id}_X(x_2) = x_2$ and therefore $f$ is injective.

• Let $x \in X.$ Then $y = f(x) \in Y$ and $g(y) = g(f(x)) = (g\circ f)(x) = \text{id}_X(x) = x$ and therefore $g$ is surjective.

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The above shows why $\boxed{f \text{ is injective and } g \text{ is surjective}}$ is true but doesn't give a reason why the other three answers aren't correct. To do that, we need to find counterexamples to the others, which is given by the following:

Set $X=[0,1], Y=[-1,1], f(x) = x, g(x) = |x|$ Then if $x\in X$ then $x\geq 0 \implies |x|=x,$ so $(g\circ f)(x) = g(f(x)) = g(x) = |x| = x = \text{id}_X(x)$ so $g\circ f = \text{id}_X$ but $g(-1) = 1 = g(1) \implies g \text{ is not injective}$ $f(x) = x \geq 0 > -1 \in Y \implies f \text{ is not surjective}$