5.1.3 Bilinear Morphisms of Pointed Sets

Let $\webleft (X,x_{0}\webright )$, $\webleft (Y,y_{0}\webright )$, and $\webleft (Z,z_{0}\webright )$ be pointed sets.

A bilinear morphism of pointed sets from $\webleft (X\times Y,\webleft (x_{0},y_{0}\webright )\webright )$ to $\webleft (Z,z_{0}\webright )$ is a map of sets

\[ f \colon X\times Y \to Z \]

that is both left bilinear and right bilinear.

In detail, a bilinear morphism of pointed sets from $\webleft (X\times Y,\webleft (x_{0},y_{0}\webright )\webright )$ to $\webleft (Z,z_{0}\webright )$ is a map of sets

\[ f \colon \webleft (X\times Y,\webleft (x_{0},y_{0}\webright )\webright ) \to \webleft (Z,z_{0}\webright ) \]

satisfying the following conditions:12

  1. Left Unital Bilinearity. The diagram

    commutes, i.e. for each $y\in Y$, we have

    \[ f\webleft (x_{0},y\webright ) = z_{0}. \]
  2. Right Unital Bilinearity. The diagram

    commutes, i.e. for each $x\in X$, we have

    \[ f\webleft (x,y_{0}\webright ) = z_{0}. \]


1Slogan: The map $f$ is bilinear if it preserves basepoints in each argument.
2Succinctly, $f$ is bilinear if we have

\begin{align*} f\webleft (x_{0},y\webright ) & = z_{0},\\ f\webleft (x,y_{0}\webright ) & = z_{0} \end{align*}

for each $x\in X$ and each $y\in Y$.

The set of bilinear morphisms of pointed sets from $\webleft (X\times Y,\webleft (x_{0},y_{0}\webright )\webright )$ to $\webleft (Z,z_{0}\webright )$ is the set $\smash {\textup{Hom}^{\otimes }_{\mathsf{Sets}_{*}}\webleft (X\times Y,Z\webright )}$ defined by

\[ \textup{Hom}^{\otimes }_{\mathsf{Sets}_{*}}\webleft (X\times Y,Z\webright ) \mathrel {\smash {\overset {\mathclap {\scriptscriptstyle \text{def}}}=}}\webleft\{ f\in \textup{Hom}_{\mathsf{Sets}}\webleft (X\times Y,Z\webright )\ \middle |\ \text{$f$ is bilinear}\webright\} . \]


Noticed something off, or have any comments? Feel free to reach out!


You can also use the contact form below: