and the source map, and their images $H_t={{\rm Im}\,}(\varepsilon_t)={{\rm Ker}\,}(H-\varepsilon_t)$ and $H_s={{\rm Im}\,}(\varepsilon_s)={{\rm Ker}\,}(H-\varepsilon_s)$ are called the target and source space. For all $g,h\in H$, we have $${\label{eq:1.1.3}}
h_{(1)}\ot \varepsilon_t(h_{(2)})=1_{(1)}h\ot 1_{(2)}~~{\rm and}~~
\varepsilon_s(h_{(1)})\ot h_{(2)}=1_{(1)}\ot h1_{(2)},$$ and $${\label{eq:1.1.3*}}
h\varepsilon_t(g)=\varepsilon(h_{(1)}g)h_{(2)}~~{\rm and}~~
\varepsilon_s(g)h=h_{(1)}\varepsilon(gh_{(2)}).$$ From [(\[eq:1.1.3\*\])]{}, it follows immediately that $${\label{eq:1.1.3a}}
\varepsilon(h\varepsilon_t(g))=\varepsilon(hg)~~{\rm and}~~
\varepsilon(\varepsilon_s(g)h)=\varepsilon(gh).$$ The source and target space can be described as follows: $$\begin{aligned}
H_t=\{h\in H~|~\Delta(h)=1_{(1)}h\ot 1_{(2)}\}=
\{\phi(1_{(1)})1_{(2)}~|~\phi\in H^*\};{\label{eq:1.1.4}}\\
H_s=\{h\in H~|~\Delta(h)=1_{(1)}\ot h1_{(2)}\}=
\{1_{(1)}\phi(1_{(2)})~|~\phi\in H^*\}.{\label{eq:1.1.5}}\end{aligned}$$ We also have $${\label{eq:1.1.6}}
\varepsilon_t(h)\varepsilon_s(k)=\varepsilon_s(k)\varepsilon_t(h),$$ and its dual property $${\label{eq:1.1.6*}}
\varepsilon_s(h_{(1)})\ot \varepsilon_t(h_{(2)})=
\varepsilon_s(h_{(2)})\ot \varepsilon_t(h_{(1)}).$$ Finally $\varepsilon_s(1)=\varepsilon_t(1)=1$, and $${\label{eq:1.1.7}}
\varepsilon_t(h)\varepsilon_t(g)=\varepsilon_t(\varepsilon_t(h)g)~~{\rm and}~~
\varepsilon_s(h)\varepsilon_s(g)=\varepsilon_s(h\varepsilon_s(g)).$$ This implies that $H_s$ and $H_t$ are subalgebras of $H$.\

[\[le:1.2\]]{} Let $H$ be a weak bialgebra over a commutative ring. Then $\Delta(1)\in H_s\ot H_t$.

Applying $H\ot \varepsilon\ot H$ to [(\[eq:1.1.1\])]{}, we find that $1_{(1)}\ot 1_{(2)}=\varepsilon_s(1_{(1)})\ot 1_{(2)}
\in H_s\ot H$ and $1_{(1)}\ot 1_{(2)}=1_{(1)}\ot \varepsilon_t(1_{(2)})\in H\ot H_t$. Now let $K_s={{\rm Ker}\,}(\varepsilon_s)$, $K_t={{\rm Ker}\,}(\varepsilon_t)$. Then $H= H_s\oplus K_s= H_t\oplus K_t,$ and $$H\ot H=H_s\ot H_t\oplus H_s\ot K_t \oplus K_s\ot H_t \oplus K_s\ot K_t,$$ so it follows that $H_s\ot H_t= H\ot H_t \cap H_s\ot H$.

The target and source map for the weak bialgebra $H^{\rm op}$ are $${\label{eq:1.2a.1}}
\ol{\varepsilon}_t(h)=\varepsilon(h1_{(1)})1_{(2)}\in H_t~~{\rm and}~~
\ol{\varepsilon}_s(h)=\varepsilon(1_{(2)}h)1_{(1)}\in H_s.$$ $\ol{\varepsilon}_t$ and $\ol{\varepsilon}_s$ are also projections.\
The source and 