# Every injection yields a bijection onto its image

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Make sure this page is up to scratch before removing this. It should be pretty good; it'd be wrong to not mark it as a stub when it was made simply to be linked to

## Statement

[ilmath]\newcommand{\f}{\overline{f}}[/ilmath]Let [ilmath]X[/ilmath] and [ilmath]Y[/ilmath] be sets and suppose [ilmath]f:X\rightarrow Y[/ilmath] is any injective map between them. Then we claim that there is a map:

• [ilmath]\overline{f}:X\rightarrow f(X)[/ilmath] given by [ilmath]\overline{f}:x\mapsto f(x)[/ilmath]- where [ilmath]f(X)[/ilmath] denotes the image of [ilmath]X[/ilmath] under [ilmath]f[/ilmath][Note 1]

that is a bijection between [ilmath]X[/ilmath] and [ilmath]f(X)[/ilmath] (that is to say, we claim that [ilmath]\f[/ilmath] is a bijection, a bijection is any map that is both injective and surjective)

## Proof

We must show that [ilmath]\f:X\rightarrow f(X)[/ilmath] is a bijection (that means [ilmath]\f[/ilmath] is both injective and surjective), even though injectiveness is (even more than surjectiveness) trivial, we do it anyway. That's why this page is marked as first-year friendly

1. Injectivity of [ilmath]\f:X\rightarrow f(X)[/ilmath] given by [ilmath]\f:x\mapsto f(x)[/ilmath]
• We will take injectivity to mean: [ilmath]\forall x_1,x_2\in X[\f(x_1)=\f(x_2)\implies x_1=x_2][/ilmath] (rather than other equivalent conditions documented on the injection page)
• Let [ilmath]x_1,x_2\in X[/ilmath] be given
• Suppose [ilmath]\f(x_1)\ne\f(x_2)[/ilmath]
• Suppose [ilmath]\f(x_1)=\f(x_2)[/ilmath]
• In order for the implication to be true, we require that this means [ilmath]x_1=x_2[/ilmath]
• Notice that [ilmath]\f(x_1)=f(x_1)[/ilmath] and [ilmath]\f(x_2)=f(x_2)[/ilmath] (by definition of [ilmath]\f[/ilmath])
• As [ilmath]f:X\rightarrow Y[/ilmath] was injective itself, this means:
• [ilmath]\forall x_1,x_2\in X[\f(x_1)=\f(x_2)\implies x_1=x_2][/ilmath]
• By hypothesis, [ilmath]\f(x_1)=\f(x_2)[/ilmath] so [ilmath]f(x_1)=f(x_2)[/ilmath]
• This means that [ilmath]x_1=x_2[/ilmath] by injectivity of [ilmath]f[/ilmath]
• So [ilmath]\f(x_1)=\f(x_2)\implies f(x_1)=f(x_2)\implies x_1=x_2[/ilmath] (then by transitivity of implies [ilmath]\f(x_1)=\f(x_2)\implies x_1=x_2[/ilmath])
• We know [ilmath]\f(x_1)=\f(x_2)[/ilmath] to be true, therefore, by definition of implies and by knowing [ilmath]\f(x_1)=\f(x_2)\implies x_1=x_2[/ilmath] we must have:
• [ilmath]x_1=x_2[/ilmath]
• This completes the first part of the proof
2. Surjectivity of [ilmath]\f:X\rightarrow f(X)[/ilmath] given by [ilmath]\f:x\mapsto f(x)[/ilmath]
• To be surjective we require: [ilmath]\forall y\in f(X)\exists x\in X[\f(x)=y][/ilmath]
• Let [ilmath]y\in f(X)[/ilmath] be given
• By definition of [ilmath]f(X)[/ilmath] (that is [ilmath]f(X):=\{y\in Y\ \vert\ \exists x\in X[f(x)=y]\}[/ilmath] remember) we see:
• [ilmath]y\in f(X)\iff\exists x\in X[f(x)=y][/ilmath]
• Choose [ilmath]x\in X[/ilmath] such that [ilmath]f(x)=y[/ilmath] (which we can do thanks to the above)
• Now [ilmath]\f(x)=f(x)[/ilmath] by definition of [ilmath]\f[/ilmath] and
• [ilmath]f(x)=y[/ilmath] by our choice of [ilmath]x[/ilmath]
• So [ilmath]\f(x)=f(x)=y[/ilmath] or just [ilmath]\f(x)=y[/ilmath]
• We have shown there exists an [ilmath]x\in X[/ilmath] such that [ilmath]\f(x)=y[/ilmath] for our given [ilmath]y[/ilmath]
• Since [ilmath]y\in f(X)[/ilmath] was arbitrary we have shown this for all [ilmath]y[/ilmath]
• That completes the surjectivity part of the proof

TODO: Check, I should have left this blank and marked it as low-hanging fruit....

Thus we have shown that [ilmath]\f:X\rightarrow f(X)[/ilmath] given by [ilmath]\f:x\mapsto f(x)[/ilmath] is a bijection