$\epsilon(\bk_{\alpha_i}) =1$ for all $i \in \I^{\imath} _{2r} $.

It follows by Proposition \[int:prop:coproduct\] that ${\U^{\imath}}_q(\mathfrak{sl}_{2r+1})$ is a (right) coideal subalgebra of $\U$. The map $\Delta : {\U^{\imath}}_q(\mathfrak{sl}_{2r+1}) \rightarrow {\U^{\imath}}_q(\mathfrak{sl}_{2r+1}) \otimes \U_q(\mathfrak{sl}_{2r+1})$ will be called the coproduct of ${\U^{\imath}}_q(\mathfrak{sl}_{2r+1})$ and $ \epsilon : {\U^{\imath}}_q(\mathfrak{sl}_{2r+1}) \rightarrow \Qq$ will be called the counit of ${\U^{\imath}}_q(\mathfrak{sl}_{2r+1})$. The coproduct $\Delta : {\U^{\imath}}_q(\mathfrak{sl}_{2r+1}) \rightarrow {\U^{\imath}}_q(\mathfrak{sl}_{2r+1}) \otimes \U_q(\mathfrak{sl}_{2r+1})$ is coassociative, i.e., $
(1 \otimes \Delta) \Delta = (\Delta \otimes 1)\Delta: {\U^{\imath}}_q(\mathfrak{sl}_{2r+1}) \rightarrow {\U^{\imath}}_q(\mathfrak{sl}_{2r+1}) \otimes \U_q(\mathfrak{sl}_{2r+1}) \otimes \U_q(\mathfrak{sl}_{2r+1})$. The counit map $\epsilon$ makes $\Qq$ a (trivial) ${\U^{\imath}}_q(\mathfrak{sl}_{2r+1})$-module. Let $m : \U_q(\mathfrak{sl}_{2r+1}) \otimes \U_q(\mathfrak{sl}_{2r+1}) \rightarrow \U_q(\mathfrak{sl}_{2r+1})$ denote the multiplication map. We have $
m (\epsilon \otimes 1)\Delta = \imath : {\U^{\imath}}_q(\mathfrak{sl}_{2r+1}) \longrightarrow \U_q(\mathfrak{sl}_{2r+1})
$ by direct computation.

The quantum symmetric pair $(\U_q(\mathfrak{sl}_{2r+2}), {\U^{\imath}}_q(\mathfrak{sl}_{2r+2}))$ {#sec:Uj}
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We set $$\label{eq:Ihf}
\I^{\imath}_{2r+1}  :=\Z_{>0} \cap \I_{2r+1} =\{1, \ldots, r\}.$$

The Dynkin diagram of type $A_{2r+1}$ together with the involution $\inv$ can be depicted as follows:

(-2,0) node [$A_{2r+1}:$]{}; (0,0) node\[below\] [$\alpha_{-r}$]{} – (2,0) node\[below\] [$\alpha_{-1}$]{} ; (2,0) – (3,0) node\[below\] [$\alpha_{0}$]{} – (4,0) node\[below\] [$\alpha_{1}$]{}; (4,0) – (6,0) node\[below\] [$\alpha_{r}$]{} ; (0,0) node (-r) [$\bullet$]{}; (2,0) node (-1) [$\bullet$]{}; (3,0) node (0) [$\bullet$]{}; (4,0) node (1) [$\bullet$]{}; (6,0) node (r) [$\bullet$]{}; (-r.north 