4.1.1 Left 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 left 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

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

satisfying the following condition:[1][2]

  • Left Unital Bilinearity. The diagram

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

    \[ f\webleft (x_{0},y\webright ) = z_{0}. \]

The set of left 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 $\smash {\textup{Hom}^{\otimes ,\mathrm{L}}_{\mathsf{Sets}_{*}}\webleft (X\times Y,Z\webright )}$ defined by

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


Footnotes

[1] Slogan: The map $f$ is left bilinear if it preserves basepoints in its first argument.
[2] Succinctly, $f$ is bilinear if we have
\[ f\webleft (x_{0},y\webright ) = z_{0} \]
for each $y\in Y$.

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