3.8 Lemmas about maps of sets

In the section \(X\), \(Y\) and \(Z\) are sets and we write \(\operatorname{\mathrm{Fun}}(X,Y) = \{ f:X\to Y\} \).

Definition 3.37

Suppose \(f:X\to Y\) is a map of sets.

We write \(f_*:\operatorname{\mathrm{Fun}}(Z,X)\to \operatorname{\mathrm{Fun}}(Z,Y)\) for the map given by \(g\mapsto f\circ g\).
We write \(f^*:\operatorname{\mathrm{Fun}}(Y,Z)\to \operatorname{\mathrm{Fun}}(X,Z)\) for the map \(g\mapsto g\circ f\).

The map \(f_*\) is sometimes called the pushforward along \(f\) and the map \(f^*\) is sometimes called the pullback along \(f\).

Proposition 3.38

Suppose \(f: X\to Y\) is a map of sets. Then \(f\) is one-one if and only if, for all sets \(Z\), \(f_*\) is one-one.

Proof ▼

(\(\Rightarrow \)) Exercise.

(\(\Leftarrow \)) Suppose that for all \(f\), \(f_*\) is one-one. Take \(Z = \{ 1\} \), the set with one element. Then \(f_*:\operatorname{\mathrm{Fun}}(Z,X)\to \operatorname{\mathrm{Fun}}(Z,Y)\) is one-one. So take \(x_1,x_2\in X\) and suppose \(f(x_1) = f(x_2)\). We can define maps \(g_i:Z\to X\) by \(g_1(1) = x_1\) and \(g_2(1) = x_2\). Then \(f_*(g_1) = f_*(g_2)\). So, as \(f_*\) is one-one, \(g_1 = g_2\). But then \(x_1 = x_2\).

Proposition 3.39

Suppose \(f:X\to Y\) is a map of sets. Then \(f\) is onto if and only if \(f^*:\operatorname{\mathrm{Fun}}(Y,Z)\to \operatorname{\mathrm{Fun}}(X,Z)\) is one-one for all sets \(Z\).

Proof ▼

Exercise.

(\(\Leftarrow \)) Suppose \(f^*:\operatorname{\mathrm{Fun}}(Y,Z)\to \operatorname{\mathrm{Fun}}(X,Z)\) is one-one for all \(Z\). I claim that \(f\) is onto.

To see this, set \(Z = \{ 0,1\} \) and define maps \(h_0, h_1:Y\to Z\) by setting \(h_0(y) = 0\) for all \(y\in Y\) and

\[ h_1(y) = \begin{cases} 0,& y\in f(X);\\ 1,& \text{else}. \end{cases} \]

Then \(f^*h_1 = f^* h_0\). So, by assumption, \(h_1 = h_0\). But then \(y\in f(X)\) for all \(y\). So \(f\) is onto.

Suppose \(f:G\to H\) is a group homomorphism. Then \(\operatorname{\mathrm{Hom}}(G,H)\subseteq \operatorname{\mathrm{Fun}}(G,H)\). If \(K\) is another group, then \(f_*(\operatorname{\mathrm{Hom}}(K,G))\subseteq \operatorname{\mathrm{Hom}}(K,H)\) and \(f^*(\operatorname{\mathrm{Hom}}(K,H))\subseteq \operatorname{\mathrm{Hom}}(G,H)\) by Corollary 3.7. So we get maps

\begin{align*} & f_*:\operatorname{\mathrm{Hom}}(K,G)\to \operatorname{\mathrm{Hom}}(K,H)\\ & f^*:\operatorname{\mathrm{Hom}}(H,K)\to \operatorname{\mathrm{Hom}}(G,K). \end{align*}

Corollary 3.42

Suppose \(f:G\to H\) is a group homomorphism. Then

\(f\) is one-one if and only if \(f_*:\operatorname{\mathrm{Hom}}(K,G)\to \operatorname{\mathrm{Hom}}(K,H)\) is one-one for all groups \(K\).
\(f\) is onto if and only if \(f^*:\operatorname{\mathrm{Hom}}(H,K)\to \operatorname{\mathrm{Hom}}(G,K)\) is one-one for all groups \(K\).

Then next goal is to determine what the image of \(f^*\) is in the case when \(f\) is onto.