A preimage of \(8\) is a real number \(x\) satisfying \(f(x)=8\). Write the equation and solve it.

The preimage of \(8\) is \(2\).

Remark: \(f(2)=8\) and the preimage of \(8\) is \(2\). Image and preimage are two opposite readings of the same equality.

Start from \(f(x)=y\) and express \(x\) in terms of \(y\). How many solutions do you get?

Let \(y\in\mathbb{R}\).

This equation has one solution and only one. Every real number therefore has exactly one preimage under \(f\).

\(f\) is affine with coefficient \(3>0\), hence increasing: the image of a segment is read off its endpoints.

\(f\) is increasing on \(\mathbb{R}\), so:

Translate \(x\in f^{-1}(]-1;5])\) into the double inequality \(-1<f(x)\leqslant 5\), then isolate \(x\).

The open and closed endpoints carry over in the same order, because \(f\) is increasing. With a decreasing function they would be swapped.

Uniqueness of the preimage gives injectivity; its existence gives surjectivity.

Exercise 1 c) establishes that every \(y\in\mathbb{R}\) has a preimage — so \(f\) is surjective — and only one — so \(f\) is injective. Consequently \(f\) is bijective, and:

Check: \(f^{-1}(8)=\dfrac{6}{3}=2\), and \(f(2)=8\).

\(1\leqslant x^2\leqslant 4\) is equivalent to \(1\leqslant |x|\leqslant 2\).

By definition, \(f^{-1}([1;4])=\{x\in\mathbb{R}\mid 1\leqslant f(x)\leqslant 4\}\), so:

The inverse image of an interval need not be an interval: here it is a union of two segments.

What is the sign of \(x^2\) for an arbitrary real number \(x\)?

For every real number \(x\), \(x^2\geqslant 0\). No real number has its image in \([-4;-1]\):

The inverse image of a non-empty set can be empty. Here the notation \(f^{-1}\) denotes the inverse image of a subset, not an inverse function: \(f\) is not bijective.

\(f\) is not monotonic on \([-2;1]\). Where is the minimum reached?

On \([-2;1]\), \(f\) decreases on \([-2;0]\) then increases on \([0;1]\). The minimum is \(f(0)=0\); at the endpoints, \(f(-2)=4\) and \(f(1)=1\), so the maximum is \(4\).

Answering \([1;4]\) is the classic mistake: reading the image from the endpoints alone is only valid for a monotonic function.

Start from \(f(x)=f(x')\) and conclude about \(x\) and \(x'\).

Let \(x,x'\in\mathbb{R}\):

\(f\) is injective.

Let \(y\in\mathbb{R}\). The equation \(2x+1=y\) has the solution \(x=\dfrac{y-1}{2}\), which is a real number. Every element of the codomain \(\mathbb{R}\) has a preimage: \(f\) is surjective.

Injective and surjective, \(f\) is bijective, with inverse:

For injectivity, look for two distinct real numbers with the same image. For surjectivity, look for a real number with no preimage.

Not injective: \(g(-1)=g(1)=1\) although \(-1\neq 1\).

Not surjective: \(-1\) has no preimage, since \(x^2\geqslant 0\) for every real number \(x\).

\(g_1\) and \(g_2\) have the same domain \([0;+\infty[\) and the same formula; only their codomain differs. Examine separately what this changes for injectivity, then for surjectivity.

Injectivity — it involves only the domain, which is the same for \(g_1\) and \(g_2\). Let \(x,x'\in[0;+\infty[\):

\(g_1\) and \(g_2\) are therefore both injective.

Surjectivity — it involves the codomain, which is what distinguishes them.

• \(g_1:[0;+\infty[\to\mathbb{R}\) — the real number \(-1\) belongs to the codomain \(\mathbb{R}\) and has no preimage in \([0;+\infty[\). \(g_1\) is not surjective, hence not bijective.
• \(g_2:[0;+\infty[\to[0;+\infty[\) — for every \(y\geqslant 0\), the real number \(x=\sqrt{y}\) belongs to \([0;+\infty[\) and satisfies \(g_2(x)=y\). \(g_2\) is surjective, hence bijective, with inverse:

\(x\in\overline{A\cap B}\) means \(x\notin A\cap B\), that is, the negation of “\(x\in A\) and \(x\in B\)”.

Let \(x\in E\).

The negation of a conjunction is the disjunction of the negations:

“\(x\notin A\) or \(x\notin B\)” means \(x\in\overline{A}\) or \(x\in\overline{B}\), that is, \(x\in\overline{A}\cup\overline{B}\). Hence:

Run the reasoning of a) backwards: every step is an equivalence.

Let \(x\in\overline{A}\cup\overline{B}\). Then \(x\notin A\) or \(x\notin B\); so it is false that \(x\) belongs both to \(A\) and to \(B\). Hence \(x\notin A\cap B\), that is, \(x\in\overline{A\cap B}\). Therefore:

The two inclusions give the equality:

Every step in a) is an equivalence: we could have concluded in one go. Double inclusion is required because it remains valid when the implications cannot be reversed — that is the method to master.

Start from \((g\circ f)(x)=(g\circ f)(x')\) and use the injectivity of \(g\) first.

Let \(x,x'\in E\) be such that \((g\circ f)(x)=(g\circ f)(x')\), that is, \(g(f(x))=g(f(x'))\).

Since \(g\) is injective: \(f(x)=f(x')\).

Since \(f\) is injective: \(x=x'\).

So \(g\circ f\) is injective.

Start from an element \(z\in G\) and work back: first through \(g\), then through \(f\).

Let \(z\in G\).

Since \(g\) is surjective, there exists \(y\in F\) such that \(g(y)=z\).

Since \(f\) is surjective, there exists \(x\in E\) such that \(f(x)=y\).

So \(g\circ f\) is surjective.

If \(f\) and \(g\) are bijective, they are injective and surjective. By a) and b), \(g\circ f\) is injective and surjective, hence bijective.

The order in which the hypotheses are used differs: one “goes down” through \(g\) then \(f\) for injectivity, and “works back” through \(g\) then \(f\) for surjectivity.