$a \colon I \to A$ we have the following. $${
 \begin{tikzpicture}[]  \PandocStartInclude{tikz/HAimplyF-1.tikz}\PandocEndInclude{input}{24}{23} \end{tikzpicture}}
    \; \stackrel{\text{(H)}}{=} \;
    {
 \begin{tikzpicture}[]  \PandocStartInclude{tikz/HAimplyF-2.tikz}\PandocEndInclude{input}{24}{23} \end{tikzpicture}}
    \; \stackrel{\text{(A)}}{=} \;
    {
 \begin{tikzpicture}[]  \PandocStartInclude{tikz/HAimplyF-3.tikz}\PandocEndInclude{input}{24}{23} \end{tikzpicture}}
    \; \stackrel{\text{(H)}}{=} \;
    {
 \begin{tikzpicture}[]  \PandocStartInclude{tikz/HAimplyF-4.tikz}\PandocEndInclude{input}{24}{23} \end{tikzpicture}}$$ Then (F) follows from monoidal well-pointedness.

Lemma \[lem:cpv\] is strengthened by the following proposition, which proves that compactness implies unitality.

\[prop:cptimpliesunital\] Any Frobenius algebra in a dagger compact category is unital.

[@carboni:matrices Remark (1) on page 503] Suppose that $\delta \colon A \to A {\ensuremath{\otimes}}A$ is a nonunital Frobenius algebra in a compact category. Define $\varepsilon \colon A
 \to I$ as follows. $$\varepsilon \;=\; {
 \begin{tikzpicture}[]  \PandocStartInclude{tikz/counit.tikz}\PandocEndInclude{input}{24}{23} \end{tikzpicture}} \;=\; {
 \begin{tikzpicture}[]  \PandocStartInclude{tikz/cptcounit.tikz}\PandocEndInclude{input}{24}{23} \end{tikzpicture}}$$ Then the following holds, where we draw the unit and counit of compactness by caps and cups (without dots). $$\begin{aligned}
     {
 \begin{tikzpicture}[]  \PandocStartInclude{tikz/cptFimpliesH-1.tikz}\PandocEndInclude{input}{24}{23} \end{tikzpicture}}
   & = 
   & {
 \begin{tikzpicture}[]  \PandocStartInclude{tikz/cptFimpliesH-2.tikz}\PandocEndInclude{input}{24}{23} \end{tikzpicture}}
     \; \stackrel{\text{(F)}}{=} \;
     {
 \begin{tikzpicture}[]  \PandocStartInclude{tikz/cptFimpliesH-3.tikz}\PandocEndInclude{input}{24}{23} \end{tikzpicture}}
     \; \stackrel{\text{(C)}}{=} \;
     {
 \begin{tikzpicture}[]  \PandocStartInclude{tikz/cptFimpliesH-4.tikz}\PandocEndInclude{input}{24}{23} \end{tikzpicture}} \\
   & \stackrel{\text{(F)}}{=} 
   & {
 \begin{tikzpicture}[]  \PandocStartInclude{tikz/cptFimpliesH-5.tikz}\PandocEndInclude{input}{24}{23} \end{tikzpicture}}
     \; = \;
     {
 \begin{tikzpicture}[]  \PandocStartInclude{tikz/cptFimpliesH-6.tikz}\PandocEndInclude{input}{24}{23} \end{tikzpicture}}
     \; \stackrel{\text{(C)}}{=} \;
     {
 \begin{tikzpicture}[]  \PandocStartInclude{tikz/cptFimpliesH-7.tikz}\PandocEndInclude{input}{24}{23} \end{tikzpicture}}
     \; \stackrel{\text{(M)}}{=} \;
     {
 \begin{tikzpicture}[]  \PandocStartInclude{tikz/cptFimpliesH-8.tikz}\PandocEndInclude{input}{24}{23} \end{tikzpicture}}
 \end{aligned}$$ That is, (U) holds.

Thus in the unital, monoidally well-pointed case, (F) and (H) are essentially equivalent. Our 